A Counterexample to Modus Ponens
Vann McGee claims that modus ponens "is not strictly valid" in an article from 1985
Curious.
I only have this example. Does anyone have more?
Opinion polls taken just before the1980 election showed the Republican Ronald Reagan decisively ahead of the Democrat Jimmy Carter, with the other Republican in the race, John Anderson, a distant third. Those apprised of the poll results believed, with good reason:
[1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.
[2] A Republican will win the election. Yet they did not have reason to believe
[3] If it's not Reagan who wins, it will be Anderson
Curious.
I only have this example. Does anyone have more?
Comments (116)
How?
1. (A ? B) ? C
2. A
3. B ? C
Which isn't modus ponens.
The conditional is formulated as if the following is true,
Republicans win then the vote ranking will be:
1. Reagan
2. Anderson
3. Jimmy Carter
It fails to consider the possibility that actually did occur:
Republicans win then the vote ranking will be:
1. Reagan
2. Jimmy Carter
3. Anderson
But (1) is true.
I thought your point was that it was supposed to be an example of an invalid modus ponens? Except it's not a modus ponens at all.
Modus ponens:
If p then q
p
therefore, q
p: A Republican wins the election,
q: If it's not Reagan who wins, it will be Anderson
So:
If A Republican wins the election, then If it's not Reagan who wins, it will be Anderson
and
A Republican wins the election
which, by MP, gives
If it's not Reagan who wins, it will be Anderson.
But if it is not Reagan who wins, it will be Jimmy Carter. So there is a prima facie case that MP reaches a false conclusion from true premises.
Sure there are other ways to pars it.
Unless you can rule out the MP parsing, showing that there are other parsings is irrelevant.
Can you rule out the MP parsing?
2. ¬(B ? C)
3. ¬A
If a Republican wins the election then if it's not Reagan who wins it will be Anderson
If it's not Reagan who wins it won't be Anderson
Therefore a Republican won't win the election
Would you say that's an invalid modus tollens? I'd say it's not modus tollens. The argument is:
1. (A ? B) ? C
2. ¬(B ? C)
3. ¬A
Weasel. Yes they did. If they "believed, with good reason" both [1] and [2], then they had deductive reason to believe [3].
Deductive not good enough? Sure. Deductive not always good enough. Too strict at times. Then go inductive.
"Strictly", though... deductively... modus ponens valid. Here, as anywhere.
Just as you can convert "A ? (B ? C)" to "(A ? B) ? C", you can also convert it back.
So " If a Republican wins the election, then if it's not Reagan who wins it will be Anderson" is logically equivalent, has the same truth-value, as "If a Republican wins the election And it is not Reagan Then it will be Anderson" (since "not" is being parsed as part of the sentence, and not an operator in the above form)
Once you convert it you have to also convert premise 2 so that the antecedent includes two conditions.
At first blush doesn't it seem like "B -> C" is false, though? Since clearly if Reagan does not win then it will be Carter. What am I missing?
[1] ... as above
[2]..... as above
[3] If it's not Reagan who wins, it will be Anderson
[4] Anderson will not win [additional background information from the scenario]
[5] Reagan will win [by modus tollens [3] and [4]]
The trick is that [1] and [2] are selected parts of information. Crucially, [4] is not mentioned. If we did not know from the scenario that Anderson was a hopeless case, we would be quite happy to accept [3] following from [1] and [2] as indeed (I submit) it does follow - by modus ponens. So modus ponens is not threatened as a logical form. Phew.
p: a Republican wins the election.
q: if it's not Reagan who wins, it will be Anderson (q = r?s)
r?s is equivalent to ¬r v s, so q should be interpreted as: either Reagan wins or Anderson wins.
Since Reagan won, q is true, since one of the components of the disjunction is true.
Before Reagan won, if it was possible that Carter could have won, they couldn't have known whether q is true or not.
But, since p is true , q is true, and p?q is true, modus ponens leads to a true conclusion from true premises.
It's just a case of the so called “paradoxes of material implication”.
P = Republicans won
R = Reagan won
A = Anderson won
P -> (~R -> A)
Let's ignore "P ->" for the moment.
Is (~R -> A) true? No!
The conditional as a whole, P -> (~R -> A) is false if P is true.
2. P
3. Therefore premise 2 is false.
Premise 2 is false= ¬p
1. p?¬p means: ¬p v ¬p, which is ¬p.
2. p
3. Therefore ¬p.
Obviously, if you have both p and ¬p as premises you can conclude anything you want, since anything follows from a contradiction.
The implication p?¬p is only true if ¬p is true. So if 2 is true, 1 is false, and you can't infer ¬p using modus ponens.
See:
Quoting Amalac
Take for instance, the following beliefs:
P (Reagan wins) = 0.80
P (Carter wins ) = 0.15
P (Andy wins ) = 0.05 (i.e. distant third republican)
P (Reagan or Andy) = 0.80 + 0.05 = 0.85 (i.e. the probability that a Republican wins)
P(Reagan | Reagan or Andy ) + P(Andy | Reagan or Andy) = 1 (i.e, as a logical tautology, Andy must win if Reagan doesn't, relative to the assumption that a republican wins)
But if Reagan doesn't win, then
P(Andy | Andy or Carter) = 0.05/ (0.05 + 0.15) = 0.25, (i.e. Carter remains favourite over Andy)
But notice that although this example contradicts (the misuse of) logical Modus Ponens, it doesn't contradict "probabilistic modus ponens" of the form P (B,A) = P( B | A) * P(A), which when summed over the values permitted for A recovers P(B).
In other words, if we take the conditional probabilities as being fundamental and follow this example in the bottom-up direction using this probabilistic modus-ponens, we recover the initial unconditional beliefs.
2.p
3. Therefore modus ponens is invalid.
'A' for 'Anderson wins'
'C' for 'Carter wins'
'R v A' for 'a Republican wins'
(R v A) -> (~R -> A)
R v A
therefore ~R -> A
That's an instance of modus ponens..
(A) If republican (x) and wins (x), then (B) (not (Reagan(x) implies Anderson (x).
(A=>B) has people in general talked about in (A), contextually only republicans are talked about in (B).
Has a different domain - it's talking about the set of presidential candidates - a republican - a republican candidate.
Is false in the overall domain {Anderson, Reagan, Carter}, but true in the domain {Anderson, Reagan}.
So while you can string match "A Republican will win the election" with making A true, it doesn't follow that string matching preserves truth condition when the strings have different implicatures . Specifically, when you move from "If" to "then" in (A), it's coming along with a domain change (from candidates to republicans).
"A republican will win the election"
Has an implicit domain of presidential candidates.
[1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.
Has an implicit domain of republicans.
Bu the puzzle includes an intensional operator "believe'.
We may state these atomic propositions purely as sentence letters so there is not a need to involve domains.
There's a set of candidates C:
{Carter, Reagan, Anderson}
[1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.
Maps {Carter, Reagan, Anderson} to {Reagan, Anderson}, the latter set of candidates is where the disjunction is understood and evaluates to true.
[2] A Republican will win the election.
Asserts that of {Carter, Reagan, Anderson}, the winner will be a member of {Reagan, Anderson}
[3] If it's not Reagan who wins, it will be Anderson
Asserts that of {Reagan, Anderson}, if Reagan doesn't win, it will be Anderson.
There are domains in the sense that the disjunctions are implicitly quantifying over those sets. That is what I meant - in the same sense that existential quantification is iterated disjunction on a finite domain.
At a point before the election, with 'wins' understood as 'will win', then R v A is true.
At a point after the election, with 'wins' understood as 'won', then R v A is true.
Quoting TonesInDeepFreeze
I was refering to Bartricks' fallacious argument, not to the OP's.
Yes, my mistake.
Yes, but the puzzle (if there truly is one) maintains with or without resort to talking about domains.
Include the premises:
R
~C
~A
Then
(R v A) -> (~R -> A)
R v A
~A
~C
therefore ~R -> A
Still valid.
Depends on the symbol interpretation. I think it's more of a problem about how the informal argument codifies into the formal logic.
In the same manner that from "You're fine" you wouldn't be able to conclude "You're pretty" or "You're okay" without knowing the context, you also wouldn't be able to assert "You're fine" is true if and only if you're fine if the quoted thing and the "disquoted" thing were from different speech events. The strings in the statement are like that, "then" in line 1 begins considering only republicans, you don't get that same subsetting effect if you assert it without "then".
Quoting bongo fury
Assume, assert, affirm, hold, "believe"... whatever.
Quoting TonesInDeepFreeze
Indeed. And perfectly valid.
If you can't stand by all 3 lines at once, don't. They can't be a good expression of what you're trying to say. Don't necessarily involve any logic in expressing yourself, but don't think you need a better one. (I don't mean you.)
'P' stands for 'Republican wins'.
'D' stands for 'Democrat wins'.
Add the background premises:
P <-> (R v A)
D <-> C
R <-> ~(A v C)
A <-> ~(R v C)
C <-> ~(R v A)
R
So, these follow:
P -> (~R -> A)
P
and conclusion:
~R -> A
/
Or spell it out with constants and predicates.
'c' stands for Carter
'r' stands for Reagan
'a' stands for Anderson
'R' stands for 'is a Republican candidate'
'D' stands for 'is a Democratic candidate'
'W' stands for 'wins the election:
Add the background premises:
Ax(Rx <-> (x = r v x = a))
Ax(Dx <-> x = c) [but not needed for the argument]
Rr
Ra
Dc [but not needed for the argument]
Wa <-> ~(Wr v Wc)
Wc <-> ~(Wr v Wa)
Wr <-> ~(Wa v Wc)
Wr [but not needed for the argument]
So, these follow:
1. Ex(Rx & Wx) -> (~Wr -> Wa) from background premises
2. Ex(Rx & Wx) from background premises
3. (Rr & Wr) v (Ra & Wa) from 2 and background premises
4. ((Rr & Wr) v (Ra & Wa)) -> (~Wr -.> Wa)
5. ~Wr -> Wa from 3,4 MP
So even with the finer analysis with constants and predicates, we still arrive at MP captured more easily anyway with just sentence letters.
So my point in response to you is that In a context of classical logic, if an argument is valid then it doesn't become invalid by adding premises of finer analysis (such as predicate logic is finer than propositional logic). This is the monotonic property of classical logic.
I essentially agree. My point is that:
(R v A) -> (~R -> A)
R v A
therefore ~R -> A
is an instance of modus ponens, but
(R v A) -> (~R -> A)
R v A
therefore we have reason to believe ~R -> A
is not.
If modus ponens is valid, then if we believe the premises, then we believe the conclusion (not always in fact - people err - but in principle). And the premises are believed but the conclusion is not. So, still, there's a puzzle.
No, it's:
(R v A) -> (~R -> A)
(R v A)
therefore ~R -> A
C <-> ~(R v A) is a given
~ R -> (C v A) is a given
C <-> ~A is a given
Lets' say:
~R -> C is a given
Then:
(R v A) -> (~R -> A)
(R v A)
therefore (~R -> A) & (~R -> C)
No contradiction.
When you believe the premises, you interpret to the string which is used to express them.
(1) If you're an apple, then you're sour or sweet or juicy.
(2) If you're not juicy, then you're sour or sweet.
(3) You're not juicy.
(4) You're either sour or sweet.
Exactly the same thing. You end up transitioning to a space of interpretations that excludes juiciness.
I take it that you intend (4) as a conclusion from the premises above it.
Quoting fdrake
I don't know what you mean by "space of interpretations that excludes". It would help if you said it in ordinary terminology for logic.
The predicates are 'is sour', 'is sweet' and 'is juicy'. I guess you mean that your intended interpretation has as its domain the set of apples?
Let 'S' stand for 'is sour', 'W' for 'is sweet' and 'J' for 'is juicy'. Let 'c' be a constant.
If the domain is intended to be the set of apples, then we don't need to symbolize 'is an apple'.
So perhaps this captures your argument:
1. Ax(Sx v Wx v Jx) premise
2. Ax(~Jx -> (Sx v Wx)) from 1
3. ~Jc premise
4 Sc v Wc from 2,3
I don't see a problem.
Or, if the intended domain is not specified, and we have 'P' for 'is an apple':
1. Ax(Px -> (Sx v Wx v Jx)) premise
2. Ax(Px -> (~Jx -> (Sx v Wx))) from 1
3. Pc
4. ~Jc premise
5. Sc v Wc from 2,3,4
I don't see a problem.
/
In my predicate argument about the election, let the intended domain be {Carter, Reagan, Anderson}.
That would make some of the background premises unneeded, but logically I don't see a problem.
Quoting TonesInDeepFreeze
I don't think the problem is that ~Jc entails (Sc v Wc) given (Jc v Sc v Wc), it's that it reads as if when one asserts ~Jc, one has established (Sc v Wc) assuming the argument is valid. Whereas in fact all that can be established is (Jc v Sc v Wc).
It couldn't've been a juicy apple!
Of course, that argument establishes its conclusion only if the premises themselves are established. I don't see a problem.
Let's see if I can make you see (how I see) the problem. How I see the problem isn't that the op is a counter example to modus ponens (I think modus ponens is valid), I see the problem as that the argument as stated can't be interpreted as a modus ponens, even though it looks like one in terms of the letters that constitute it - how it's written.
I have a box, it contains an apple, an orange, or a banana, you don't know which. But you do know it can only be one of those three. It can't be more than one of those three either. It contains exactly one fruit item.
One of {apple, orange, banana} will be picked out.
Analogously:
One of {Reagan, Carter, Anderson} will win.
Let's say you believe that (assume that, posit, assume as a premise) you will pick out round-ish fruit (apple or orange). In that space of assumptions, (apple, orange), if it's not an orange it must be an apple. not(apple) implies orange holds in that domain, because it consists only of an apple and an orange.
Similarly, Republican consists only of Anderson and Reagan.
If you wanted to conclude that you will receive an apple if you don't receive an orange, you would need to eliminate the possibility of receiving a banana. You can't do that.
What you can do is eliminate the possibility of receiving a banana if you have already assumed, or it is true that you will have received, a roundish fruit. That follows from the assumption. But they can't exclude the banana, so they have no reason to believe (in the OP's terms) that they wouldn't receive a banana (analogously, a democrat, Carter, would win).
So when the words go into your eyeballs, despite the literal characters tracing out a clear instance of modus ponens, as Banno wrote:
Quoting Banno
The overall interpretation is different from what you would expect - writing it like:
Has (C) talking about all the candidates, it's evaluated over the candidates - it could be Carter.
But when you read:
The context of "it's" in the "then" part of the if-then references only Republicans. A context, current Republican candidates, in which the disjunction between Reagan and Anderson (Reagan or Anderson will win) holds.
I'm thinking that your notion of context is along the right track.
The Op was a request for further on the article. There's a trail of academic articles, various detailed accounts.
But I'm going to go with the MP being valid, and hence the conclusion true; Given that a republican won, it is true that If it's not Reagan who wins, it will be Anderson. To get to "it is true that If it's not Reagan who wins, it will be Carter", an additional premiss is needed.
Not if you hear it, for no reasons that are obvious to me, as talking about psychology. I hear it, for reasons of charity and extensionalism, as dialect for "therefore we have deductive reason to assert"... i.e. "therefore".
That still breaks the form of modus ponens.
'we have deductive reasons to assert is' is intensional.
If it were merely a flourishing touch, then we could delete it, but if we delete it then then the puzzle fizzles in the form it's given, as its form is not modus ponens if it injects a modal operator.
If P then Q
P
Therefore Q
is modus ponens.
If P then Q
P
Therefore [modal operator] Q
is not modus ponens.
Quoting bongo fury
On the contrary, it introduces intensionality.
That is logically equivalent to “Reagan or Anderson wins the election”.
Given that a Republican wins the election, it is valid to conclude that Reagan or Anderson wins the election.
If Reagan wins the election, it’s true that Reagan or Anderson wins the election.
So if they have good reason to think that a Republican will win the election because they have good reason to think Reagan will win the election, then they also have good reason to think that Reagan or Anderson will win, or equivalently, if not Reagan then Anderson. Because in believing that a Republican will win, they’ve ruled out Carter already.
Of course, but that doesn't address the puzzle.
The argument as stated is not modus ponens. It injects a modal operator in front of the conclusion.
But there is still a puzzle:
Quoting TonesInDeepFreeze
Maybe I'll get time to analyze the rest of your post.
The conclusion is not valid. The conclusion is contingent.
The modus ponens argument
(R v A) -> (~R -> A)
R v A
therefore ~R -> A
is valid.
But the argument is not of that form. It's of this form:
(R v A) -> (~R -> A)
R v A
therefore [modal operator] ~R -> A
But there is still a puzzle, as I mentioned.
I think you're erroneously reading in a modal operator (and which one are you reading in?)
The claim is that the proposition
[1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.
and the proposition that
[2] A Republican will win the election.
do not together give good reason to believe that
[3] If it's not Reagan who wins, it will be Anderson.
That "do not together give good reason to believe" is just the writer denying the validity of the argument; modals about belief are not part of the structure of the argument itself.
Since if it is given that a Republican will win the election, and that the only Republican alternative to Reagan is Anderson, it is valid to conclude that if not Reagan then Anderson will win the election, either there mustn't actually be good reason to believe at least one of those premises, or there is in fact good reason to believe the conclusion, contrary to the writer's claim.
I see your point. But I haven't been in disagreement.
I don't dispute the author's argument about the modus ponens argument.
My point is to be careful not to take his example in the form he literally gave it.
P -> Q
P
Therefore [modal]Q.
Rather that his analysis can be stated along the lines of:
Quoting TonesInDeepFreeze
This does not prove the invalidity of modus ponens. Rather it shows that modus ponens may fail our expectations of belief. And it does seem to me to be a genuine puzzle.
I think we are in agreement.
Quoting Pfhorrest
Expressing it as a different propositional syllogism, introducing predicates, and introducing modality are all unjustified and overly complicated.
We deal with it by assuming MP is correct. So we need to look at what the consequence is for our beliefs. The syllogism says nothing about Carter, so no conclusions about Carter can be reached. The only other contender is Anderson, so the conclusion has to be about Anderson or Reagan. Given that, it is true.
The discussion fo domains is close.
In case it matters, we note that the text mentions both 'good reason to believe' and 'reason to believe'.
1. we have good reason to believe (R v A) -> (~R -> A)
2. we have good reason to believe R v A
therefore 3. we have reason to believe ~R -> A
That does not prove the invalidity of modus ponens. But it is a puzzle.
Mentioning both 'reason to believe' and 'good reason to believe' suggests degrees of reasons to believe. Or perhaps the author didn't mean to imply degrees. In that case we have one of these two:
1. we have good reason to believe (R v A) -> (~R -> A)
2. we have good reason to believe R v A
therefore 3. we have good reason to believe ~R -> A
That seems to preserve the puzzle.
1. we have reason to believe (R v A) -> (~R -> A)
2. we have reason to believe R v A
therefore 3. we have reason to believe ~R -> A
That doesn't seem as strong a puzzle, but still a puzzle.
Or take the modal operator outside the scope of the argument itself and we have three versions. Of the three, the first is closest to the author's text:
1. (R v A) -> (~R -> A)
2. R v A
therefore 3. ~R -> A
we have good reason to believe 1.
we have good reason to believe 2.
we believe that modus ponens is valid, so we have reason to believe 3.
or
1. (R v A) -> (~R -> A)
2. R v A
therefore 3. ~R -> A
we have good reason to believe 1.
we have good reason to believe 2.
we believe that modus ponens is valid, so we have good reason to believe 3.
1. (R v A) -> (~R -> A)
2. R v A
therefore 3. ~R -> A
we have reason to believe 1.
we have reason to believe 2.
we believe that modus ponens is valid, so we have reason to believe 3.
But if it's just 'believes' then there is a chink in the armor:
1. (R v A) -> (~R -> A)
2. R v A
therefore 3. ~R -> A
we believe 1.
we believe 2.
we believe that modus ponens is valid, so we believe 3.
If someone claimed that they believe 1 and 2, but not 3, then I might say, "I don't think you really do believe 2."
Change to 'knows', and the puzzle is even weaker:
1. (R v A) -> (~R -> A)
2. R v A
therefore 3. ~R -> A
we know 1.
we know 2.
so we know 3.
If someone said they know 1 and 2 but not 3, then I might say, "Then wake up and smell the coffee: you're just not following through to accept knowledge implied by what you do know."
So I think the specific nature of the intensionality does have something to do with this puzzle.
Quoting fdrake
The domain is {apple, orange, banana}.
{apple, orange} is a subset of the domain. {apple orange"} is not a "space of assumptions". It is not a set of assumptions. It's a set whose members are two different pieces of fruit.
Quoting fdrake
If x is in {apple orange} and x is not the orange, then x is the apple. And if x is in {apple orange} and x is not the apple then x is the orange.
Quoting fdrake
No, we can reason from an assumption that you won't receive the banana. That the banana is in the domain doesn't entail that we can't assume that you won't receive the banana. Let 'W' stand for 'we receive x'.
The domain is {apple orange banana} but that doesn't stop us from reasoning from the premise:
W(apple) or W(orange)
Let the domain = {0 1 2}. Let 'W' stand for 'x is the number of pens' in my pocket.
Suppose I know I have a pen in my pocket, so I have the premise:
W(1) or W(2)
Or the author's example:
Let the domain be {Reagan, Anderson, Carter}. (By the way, Anderson ran as an Independent, not as a Republican, though he was a Republican.)
Let 'W' stand for 'x wins'.
Then we have the premise:
W(Reagan) or W(Anderson)
Quoting fdrake
Yes, just as we assume a Republican will win.
Quoting fdrake
No, they have very good reason: the polls. But it doesn't matter about the factual givens anyway. For sake of argument we accept that we have good reason to believe that a Republican will win and moreover that we assume a Republican will win.
Quoting fdrake
I already answered that.
Again, we can take it merely propositionally.
R for 'Reagan wins'
A for 'Anderson wins'
C for 'Carter wins'
(R v A) -> (~R -> A)
R v A
therefore ~R -> A
But why did we adopt "R v A"? Because Reagan looked bound to win. So we got it from the theorem
R -> (R v A)
So maybe it's not modus ponens that should be in question, but "R -> (R v A)".
I'm not saying we should doubt the validity of "R -> (R v A)". But maybe it's the one not mixing well with intensionality and not so much podus ponens. I think that might be right. Because we can can do it this way, without modus ponens:
1. R
2. R v A
therefore 3. ~R -> A
I think there is something to what you say. But I don't know whether we need the notion of domains for it.
She has good reason to believe she will receive the apple.
She believes that (A -> (A v O)) is valid.
So she has good reason to believe she will receive the apple or she will receive the orange.
She believes that (A v O) -> (~A -> O) is valid
So she has good reason to believe that if she doesn't receive the apple then she will receive the orange.
But she doesn't have good reason to believe that if she doesn't receive the apple then she will receive the orange.
And that's a puzzle.
But why doesn't she have good reason to believe that if she doesn't receive the apple then she will receive the orange? Because she has good reason to believe that if she doesn't receive the apple then she will receive the banana. (That's where your line of thinking comes in.)
So the banana comes up regarding her beliefs, but it doesn't come up in the argument itself.
So how can that be used to solve the puzzle?
This makes me want to abandon my suggestion that maybe its more about disjunction and intensionality than about modus ponens and intensionality. Maybe it's something about deduction and intensionality in genera (or maybe even more generally about inference and intensionality in general?)
The quote in the OP is not just about whether someone believes things, but about whether they have good reason to believe them. That's also what logical inferences (like modus ponens) are all about: it's not that believing the premises entails believing the conclusion, but that believing the premises gives reason to believe the premises. As you say, people err, and sometimes don't believe what they have reason to. But the quote in the OP is claiming that having good reason to believe the premises doesn't constitute having good reason to believe the conclusion. That's incorrect: having good reason to believe the premises does constitute having good reason to believe the conclusion.
If you have good reason to believe that:
[1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.
(say, because those are the top two highest-polling candidates in the Republican primary)
and you have good reason to believe that:
[2] A Republican will win the election.
(say, because Reagan is the top-polling candidate in the general election, so you reasonably think that Reagan, a Republican, will win)
then you do have good reason to believe that:
[3] If it's not Reagan who wins, it will be Anderson
(because that's logically equivalent to "Reagan or Anderson will win", and either "Reagan wins" or "Anderson wins" satisfies that, so if you have, say, good reason to believe that Reagan will win, which is your reason for believing [2], then you also have good reason to believe [3]; or if you have good reason to believe that Anderson will win, same thing; or if you don't have good reason to choose between Reagan and Anderson but you have good reason to expect Carter to lose; anything that gives you reason to believe [2] also gives you reason to believe [3], so long as you've also got some reason to believe [1]).
Yes, and I took account of that in followup posts. Actually, he mentions both 'good reason to believe' and 'reason to believe'. I would guess he didn't mean that difference as playing a role, but it might be good to see what happens with the distinction and without the distinction.
Quoting Pfhorrest
"all about" is sweeping. I don't take logical inference to be "ALL about" [all-caps added] good reasons for belief. Logical inference can take place in a machine that doesn't even have beliefs or reasons for belief. Modus ponens and other deductive forms have settings other than grounds for belief.
Most of the rest of your post is explanation of what I understood when I read the first post in this thread. That's okay though, as other readers may benefit from it.
He said there is good reason to believe the premises, but not a reason to believe the conclusion. And that is true*. The part about "constituting" or anything like it, is not in the quote. We might think it is fair to think he intended that (I don't know; it's a fine point); but he didn't actually say it.
* It's true given the background premise that the polls showed Reagan far ahead. But that premise is false, since Reagan was not far ahead in the polls. No matter for the analysis though, as we may take it hypothetically that Reagan was far ahead or that we had good reason to believe he would win on any other grounds.
The example also highlights a general problem: given a state of knowledge, is it consistent? and if so, how do you determine what the underlying arrows are?
In the previous example of the OP, the beliefs given are consistent. The arrows are the conditional probabilities of candidates winning given knowledge of the failure of one or more of the remaining candidates, and there is only one premise, namely that a republican wins.
To emphasize that point. The validity of modus ponens bears upon grounds for belief, but the validity of modus ponens can be (and often is) understood irrespective of grounds for belief. The validity of modus ponens is that if the premises are true then the conclusion is true, And that "the premises are true then the conclusion is true" is true of modus ponens no matter whether we even wish to raise the subject of grounds for belief.
Inferences may be drawn irrespective of belief. One could draw inferences from modus ponens all day long without even giving a thought as to what one thought are grounds for belief of anything. Indeed, in a formal sense, an argument is an ordered pair
where P is a set of sentences (or formulas but that complicates this with a technicality) and c is a sentence (or formula). A valid argument is an argument such that in all models in which all the members of P are true are models in which c is true. Then sound systems of logic are ones such that proofs only result in valid inferences. And a valid inference is one such that, again, in all models in which all the members of P are true are models in which c is true. There is no requirement that we mention "reason to believe" or anything like that. So inference isn't "ALL about" reasons for belief.
But it’s not, which is my point. If there is good reason to believe those premises, then there is reason (even good reason) to believe the conclusion.
As I elaborated, whatever reason there is to believe the second premise (R leads in polls, A leads in polls, C trails in polls, etc) is also a reason to believe the conclusion (which is true if either R or A wins), provided the first premise is also supported.
That seems right, of course. But from a different view, there is not a good reason to believe the conclusion, since there is an overwhelming better reason to believe that if Reagan does not win, then Carter wins, so that Anderson does not win. That there is both good reason to believe the conclusion and not good reason to believe the conclusion is the paradox.
Probability theory, which is currently the most fashionable calculus for representing and reasoning about beliefs and uncertainty, is only defined up to a measure over a sigma-algebra of sets denoting a collection of propositions. Unfortunately, practitioners of the theory don't normally consider this collection to be a model of any specific set of logical axioms, but rather as representing classes of observables, which means that modus ponens is formally absent from probability theory. Whenever an underlying logical system isn't specified in an application of probability theory (which is nearly all of the time), it is undetermined as to whether conditional probabilities or joint probabilities are the more fundamental epistemic principle.
Nevertheless, it is natural for Bayesian practitioners to assume some implicit underlying logic in an ad hoc fashion and to interpret modus ponens in terms of set intersections, in Venn diagram fashion. But as the example demonstrates, probabilities can behave non-intuitively with respect to modus ponens. Formally, Modus ponens speaks only of logical possibilities and not probabilities which are property of a model of a logic.
I'm not inclined to quibble with the givens of the problem or appeal to lack of certainty. That seems not to face the structure of the problem head on.
I guess we could say that there is good reason to believe the conclusion and that there is good reason not to believe the conclusion. Which in its form is not a contradiction.
The reason for believing that the conclusion is false is a good reason. So maybe its a better reason than the reason for believing the conclusion is true. So maybe its such a better reason that it makes the reason for believing the conclusion is true really not a good reason. But the reason for believing the conclusion is true is that it follows from a sound argument (true premises and modus ponens), and you can't get a better reason than that! Thus, still a puzzle.
There might be something lurking in the notion of 'good reason' that has to do with degrees of good reason, which also relates to degrees of confidence in beliefs. And Pfhorrest broaches the matter of lack of certainty. I'm not inclined to it, but maybe a solution does lie in that direction.
In logic, either an arrow A -> B exists, or it does not. And so for logic there exists only possibility or non-possibility. On the other hand, probability measures over a set of propositions in a model of logic are chosen freely in accordance with external beliefs or experiments.
On the left side below are the axioms of OP's problem that specify every possible election outcome. On the right side is an example of a consistent set of degrees of confidence assigned to each possibility that coheres with every premise of the OP.
Andy or Carter --> Andy 0.25
Andy or Carter --> Carter 0.75
Reagan 0.80
Carter 0.15
Andy 0.05
As usual, Modus Ponens holds while saying nothing about the relative likelihood of possible winners.
After all, it's not like we have parentheses designating which conditional to evaluate first.
It could just be a matter of bad translation.
I don't think I quite grasped the argument before, but I think I get it now. Am wondering if there are other nested conditionals that have a (on the surface) false conditional as its consequent, with a true premise...
Yes, I'd say this.
I'm saying that the nested conditional in logic does not behave like a string of two if-then statements in English -- so it's not a matter of applying rules of inference to the way premise 1 is set out, but trying to find a different, reasonable interpretation of the English sentence into a logical syntax that keeps MP intact.
Anderson ran as Independent, but he was a Republican. It doesn't matter anyway, since we don't need to mention 'Republican', as we could just say 'Reagan or Anderson'. Moreover, we could say 'Reagan or x' for any x whatsoever. We could say:
If either Reagan wins or Donald Duck wins, then if Reagan doesn't win then Donald Duck wins.
Either Reagan wins or Donald Duck wins.
Therefore, If Reagan doesn't win then Donald Duck wins.
or
If either Reagan wins or Carter wins, then if Reagan doesn't win then Carter wins.
Either Reagan wins or Carter wins.
Therefore, If Reagan doesn't win, then Carter wins.
But with that argument, there's no puzzle.
/
The actual factual error in the problem is the claim that Reagan was way in the polls. Actually the polls were close between Reagan and Carter.
Yes, and neither is an example of modus ponens. The syntax of the problematic version simply gives the false appearance of one. Much like a sentence such as "this sentence is false" gives the false appearance of a truth-apt proposition.
As explained here "logical forms are semantic, not syntactic constructs", and the semantics of "A ? (B ? C)" isn't the same as the semantics of "A ? D". You can't just substitute D for B ? C and have the same logical form.
That's what I did?
First, we should put aside quibbles about (a) Anderson running as Independent and (b) the mistaken claim that Reagan was far ahead in the polls. We should just take the problem at face value and take as stipulated the hypotheses that 'Republican' includes both Reagan and Anderson and that Reagan was far ahead in the polls.
/
The validity of modus ponens is:
(a) When the premises are true then the conclusion is true.
The validity of modus ponens is not:
(b) When there is good reason to believe the premises are true then there is good reason to believe the conclusion is true.
So I don't think the example belies the validity of modus ponens.
But we might claim that if (b) fails then modus ponens is not reliable for informing our belief, but we do expect that modus pones is reliable for informing our belief, as indeed we have not just good reason, but irrefragable reason, to believe modus ponens is reliable for informing our belief. So it is a puzzle.
/
McGee actually wrote not simply about good reason for belief, but about was in fact believed. His argument can be fairly paraphrased:
(1) People believed and had good reason to believe: If either Reagan wins or Anderson wins, then if Reagan does not win then Anderson wins.
(2) People believed and had good reason to believe: Either Reagan wins or Anderson wins.
(3) People did not have reason to believe: If Reagan doesn't win then Anderson wins.
We can add (a) if people did not have reason to believe, then, a fortiori, they did not have good reason to believe, and (b) other than unjustifiably optimistic Anderson supporters, people did not believe that if Reagan doesn't win then Anderson wins.
But I don't know whether the particular wording changes the puzzle.
/
'has reason to believe' and 'has good reason to believe' are intensional:
Suppose there is a spy who stole documents from Interpol and that Smith is that spy. And Jones knows about the caper but little of its details. Then:
Jones has good reason to believe "the spy is the spy". But Jones does not have good reason to believe "Smith is the spy".
If we take out 'has good reason to believe' and leave only 'believed' then we have:
(4) People believed: If either Reagan wins or Anderson wins, then if Reagan does not win then Anderson wins.
(5) People believed: Either Reagan wins or Anderson wins.
(6) People did not believe: Reagan doesn't win then Anderson wins.
That mentions belief, but intensionality is not present. It is just three statements about what people believed.
(4) and (5) are quite unlikely true if by 'people' we mean typical people, even typical people well informed about the campaign, even just journalists and political scientists. Such people never had such thoughts as "If either Reagan wins or Anderson wins, then if Reagan does not win then Anderson wins" and "Either Reagan wins or Anderson wins"*.
* There it does matter that we say "A Republican wins" rather than "Reagan wins or Anderson wins", since the former was believed.
But we should be generous to McGee by revising to (a) "People who were informed about the campaigns and understood formal logic and were presented with such a proposition would have believed". Then there is no harm in taking "People believed" to stand for (a). And then (4) and (5) are true.
All that is shown in this version is that people believed certain premises but not a conclusion that follows from those premises. That's just a factual matter. It doesn't belie modus ponens.
/
Does temporality bear on the puzzle?
McGee's version uses future tense. 'wins' stands for 'will win'. Keeping consistent tense:
(7) People believed and had good reason to believe: If either Reagan will win or Anderson will win, then if Reagan will not win then Anderson will win.
(8) People believed and had good reason to believe: Either Reagan will or Anderson will win.
(9) People did not believe* and did not have reason to believe: If Reagan will not win then Anderson will win.
* I added 'did not believe' because it is true and fits the the pattern.
Recast in past tense, and hypothesize that the people mentioned lost access to information about the election starting with news about the voting results:
(10) People believed and had good reason to believe: If either Reagan won or Anderson won, then if Reagan didn't win then Anderson won.
(11) People believed and had good reason to believe: Either Reagan won or Anderson won.
(12) People did not believe and did not have reason to believe: If Reagan didn't win then Anderson won.
Still obtains as a puzzle. So I don't see temporality as bearing on the puzzle.
/
I mentioned "R -> (R v A))". Most people don't believe it, since they don't even know about, but they would believe it if they knew about formal logic, so they do have good reason to believe it.
Another poster mentioned material implication with its clause "'False antecedent then false consequent' is true". I bet most people don't believe that, since they've never heard of it, and they wouldn't believe it even if they knew about it. And a lot of people who have heard of it don't buy it. But there are people who do believe it, especially if they accept the first chapter in a logic book, so we can limit to those people.
Here's one (which reflects the election example). Suppose I have a 3-sided die with the following roll statistics:
Before the die is rolled, I would have reason to believe 1 would be rolled (since it is rolled 80% of the time). If not 1, then I would not have reason to believe 3 would be rolled (since, when 1 is not rolled, 2 is rolled 95% of the time while 3 is only rolled 5% of the time).
Note: "It" below refers to the upward face of the rolled die.
(A) If it's odd then if it's not 1 then it's 3 [per the die characteristics]
(B) It's odd [premise]
(C) If it's not 1 then it's 3 [from (B),(A)]
Note that (B) eliminates face 2 as a possibility. Given that context, (C) can be interpreted as applying to just the odd die faces, in which case the inference is valid. In the broader context of all the die faces, the inference would be invalid (since 2 is a possibility and, furthermore, the possibility I would have reason to believe if 1 has been eliminated).
I gather that is the general point that @fdrake has been making about the domain change and the problem of how to formally capture the informal argument.
My question is, how would the argument be formally written to express that context change, and would that argument involve modus ponens or not?
:up:
Logic specifies what can happen, but not what will happen. After all, if that weren't the case, then an axiomatic system such as Peano arithmetic wouldn't be a forest of proofs, but merely a single proof of one result consisting of a single chain of reasoning.
Needless to say, there is an (unfortunate) temptation among philosophers and mathematicians to mix the concepts of logic/possibility with statistics/probability by considering conditional-probabilities to be a generalised form of logical implication. This is generally disastrous, because possibilities are easier to state and justify than probabilities which are usually ill-defined and whose use is generally controversial.
That's really good. It puts the puzzle in stark formal terms and takes out the background noise about the historical election facts. Thanks.
Quoting Andrew M
I don't get that. The logic is monotonic. So how can adding premises make the argument invalid? And how would we formalize the inclusion of a broader context? I surely see the point that not mentioning (2) relates to the problem, but I don't know how we would formulate that other than just mentioning it, and how it would overturn an argument in a monotonic logic.
Meanwhile, I'm inclined to think that a solution would center around problems with the notion of "good reason to believe".
The solution is that it is not puzzling, let alone paradoxical, to have good reason to believe a statement and also good reason to believe the negation of that statement. Happens all the time in life when we are confronted with a tough decision.
Restating the problem:
Suppose there are three candidates in an election: R, A, and C.
Suppose, the day before the election, the polls show 60% for R, 5% for A, and 35% for C.
Let Jack be a person who knows those poll numbers but he slept through the election so he doesn't know who won.
(1) (R v A) -> (~R -> A)
(2) R v A
Therefore, (3) ~R -> A
Jack has good reason to believe (1).
Jack has good reason to believe (2).
Jack has good reason to believe (3).
Jack has good reason to believe (4) ~R -> C.
There is not a contradiction there.
To get a contradiction, we need to derive:
(5) Jack does not have good reason to believe (3)
But it wouldn't be by logic alone, since
~R -> A
~R -> C
~(C & A)
is a consistent set as seen by this model (which happens to be the real world):
R is true
C is false
A is false
Yes, we do get ~(~R -> A).
But we haven't yet derived a contradiction about Jack's good reason for belief. So the burden is on McGhee to show a contradiction, especially since it is not puzzling, let alone paradoxical, that one has good reason to believe a statement and also good reason to believe the negation of that statement.
But still, it does stick in the craw to say "Jack has good reason to believe that if R didn't win then A won".
/
But what if we changed the argument to this:
(1) (R v A) -> (~R -> A)
(2) R v A
Therefore, (3) ~R -> A
Jack has good reason to believe (1).
Jack has good reason to believe (2).
Jack has good reason to believe (1) and (2) imply (3).
So Jack has good reason to believe (3).
Jack has good reason to believe (4) ~R -> C.
I don't see that it changes anything materially.
/
Also, if 'good' is not being used in the analysis, then we can drop it, and just say 'reason to believe'.
Precisely. the proposition ~R --> A isn't in contradiction with the proposition ~R -->C because both denote possibilities, as opposed to probabilities or propensities. To get the latter, a non-logical probability measure must be added.
Or alternatively, since precise probabilities are usually difficult and controversial to assign, one simply ranks ~R --> C above ~R --> A to indicate which they believe is the most likely.
:up:
Quoting TonesInDeepFreeze
By broader context, I meant a context where we consider only the characteristics of the die where face 1, 2 and 3 are all possibilities. So we might say, "If it's not 1 then it's 2". That's not a valid inference (since 3 is also remotely possible), but it's a reasonable belief based on the stated probabilities.
Whereas the more specific context includes (B) which eliminates face 2 as a possibility. So in that context we might say "If it's not 1 then it's 3" which is a valid inference and also a reasonable belief (since there are no other possibilities).
Quoting TonesInDeepFreeze
I think so as well. Initially (based on the polls), there's good reason to believe that if Reagan doesn't win then Carter will. But it's not a valid inference, since there is a remote possibility that Anderson will win.
When we subsequently learn that a Republican has won (or will win), then there is no longer good reason to believe that if Reagan doesn't win then Carter will, since Carter has been eliminated as a possibility. So the remote possibility of Anderson winning becomes the only possible alternative to Reagan winning. So there is now good reason to believe that if Reagan doesn't win then Anderson will. It's a valid inference, even though Anderson winning remains only a remote possibility.
Right, I understood that.
Quoting Andrew M
I don't understand that.
Right, ~1 > 2 is not entailed when there is not a premise 1 v 2. But the reason it is not entailed is just logic. I don't see what the possibility of 3 has to do with.
Maybe you meant that the possibility of 2 should allow ~1 -> 2 as a possibility?
But 'possibility' is bringing a modal operator.
The premises are purely sentential:
[original argument:]
(background assumption) 1
(from background assumption) 1 -> (1 v 3)
(A) (1 v 3) -> (~1 -> 3)
(B) 1 v 3
therefore (C) ~1 -> 3
I do see this:
[revised argument:]
(background assumption) 1
(from background assumption) 1 -> (1 v 2 v 3)
(A') (1 v 2 v 3) -> (~1 -> (2 v3))
(B') 1 v 2 v 3
therefore (C) ~1 -> 3 WRONG
But that doesn't make the original argument incorrect.
Quoting Andrew M
It is valid from the background assumption that Reason wins.
"Reagan wins" is how we got "a Republican wins", which means "Reagan wins or Anderson wins".
Both ~R -> A and ~R -> C are entailed from the background assumption that Reagan wins.
But ~R -> C is not entailed from just "a Republican wins" which is R v A.
And of course, that is consistent.
So my solution is that there is good reason to believe both ~R -> A and ~R -> C.
Though it is counterintuitive to believe ~R -> A.
So there is good reason to believe something that is counterintuitive. And that is counterintuitive. (Is it paradoxical?) And modus ponens ponens is not invalid. And I think the problem has more to do with disjunction than with modus ponens. That aligns with you and fdrake in the sense that the puzzle results from leaving off Carter in the disjunction.
I see your point.
A logic form may not be comprehensive. A simple example:
Let P = AxRx
Let Q = Ra
P
therefore Q
INVALID
AxRx
Ra
VALID
https://thephilosophyforum.com/discussion/comment/567916
I think we're interpreting the problem differently. You regard 1 as the background assumption, whereas I regard (1 v 2 v 3) as the background assumption (i.e., the die can roll 1, 2 or 3).
What a person has good reason to believe (if not 1 then 2) is distinct from the logic of the situation (1 v 2 v 3).
When the person learns that an odd number has been rolled, then the logic becomes (1 v 3). That knowledge update is a change of context, and the person's reasoning changes. They now have good reason to believe (if not 1 then 3), since 3 is now the only possible alternative to 1, albeit remote.
Quoting TonesInDeepFreeze
As I interpret the situation, ~R -> A is not counterintuitive when derived in the appropriate context. Given the polls, a person has good reason to believe a Republican has won (or will win). But Carter might still have won, despite their good reason, since their good reason is not sufficient for truth.
On the other hand, a person could learn that a Republican has won. Given their updated knowledge (a change of context), Carter cannot have won since Carter is not a Republican. So, given their newly acquired knowledge, if Reagan didn't win, then Anderson did.
So I think that interpretation leaves modus ponens as valid and also shows how ~R -> A can be intuitive when derived in the appropriate context.
I was mistaken to couch it the way I did.
Now I don't think ~R -> A is counterintuitive. Because it has probability of 65%
Exactly.
I realized that this has an even simpler explanation.
There really isn't a puzzle.
It has not been shown by McGee that modus ponens does not preserve strength for reason to believe.
And the point I mentioned about best strength is not relevant either.
And it has nothing more to do with modus ponens than with tautology itself.
This is so simple that I can't believe I didn't see it.
The only non-logical premise is R v A.
And the conclusion ~R -> A is equivalent with R v A.
So it's just a tautological inference.
The strength of R v A is 65%. And that strength is preserved from the non-logical premise to the non-logical conclusion.
Suppose the only non-logical premise is R v C.
The conclusion ~R -> C is equivalent with R v C.
So it's just a tautological inference.
The strength of R v C is 95%. And that strength is preserved from the non-logical premise to the non-logical conclusion.
That's all. Two different non-logical premises in two different arguments, and their strength preserved in the conclusion in both arguments.
/
For reference, here's the setup of the problem:
Suppose there are three candidates in an election: R, A, and C.
Suppose, the day before the election, the polls show 60% for R, 5% for A, and 35% for C.
Let Jack be a person who knows those poll numbers but he slept through the election so he doesn't know who won.
(1) (R v A) -> (~R -> A)
(2) R v A
Therefore, (3) ~R -> A
If p then q,
not q, (because neither Reagan nor Anderson won)
therefore not p
That's modus tollens though.
If you want to stay in the MP, it should be:
If p then q,
not p, (because no Republican won)
therefore not q.
The MP is perfectly valid.
It seems more like slight of hand to play with the implied meaning of Reagan being a Republican and leaving out part of q because that's "either Reagan or Anderson wins" and not only "Reagan wins".
I think fdrake and Andrew M had the right idea, but it needed a follow-through. I think sime had the solution in a general form.
I found this problem extremely interesting because in human inference making modus ponens is about as basic and ubiquitous an argument form there is, so it would be astounding to find that modus ponens is not reliable.
I can't find McGee's article on the Internet. If there is more in the article that materially qualifies the clip in the first post of this thread, then my remarks might need to be modified. But at this time I'm taking the clip at face value along with the quote "is not strictly valid".
If McGee meant this as a joke or magic trick, then I would say it is a very clever and entertaining joke or magic trick. But I take it that it was meant seriously, so I am curious why he didn't himself see the fallacies in his argument. It turns out that, when you unpack his argument, the solution to his challenge is trivial. So it is fun to see a baffling problem turn out to have a trivial solution.
.
I use a hypothetical person named 'Jack' instead of a group of people referred to as 'they.
I take McGee's argument to be fairly couched this way :
We start with premises that Jack has good reason to believe, then we arrive at a conclusion that (a) Jack does not believe and (b) doesn't have good reason to believe. Therefore, modus ponens fails to preserve strength of reason for belief. Therefore, modus ponens is not strictly valid.
That depends on the assumption:
For modus ponens to be strictly valid, modus ponens must preserve strength of reason for belief.
We should accept that assumption, at least for sake of argument.
But (a) is irrelevant. Argument forms don't ensure that people believe the conclusions. People err in their beliefs; that's not the fault of argument forms.
As for (b), argument forms pertain to what is the case, or (granting McGee's assumption) what should be believed to be the case, only relative to the premises. And relative to the premises, it is not the case that there is not good reason to believe the conclusion ~R -> A.
I assigned specific probabilities. But they could be any probabilities, as long as they entail that there is good reason to believe R v A. So here's the example with unspecified probabilities.
prob(R & A) = 0
prob (R & C) = 0
Assume prob(R v A) is great enough that we have good reason to believe R v A.
Assume prob(R v C) > prob(R v A).
The only non-logical premise in the modus ponens is R v A. And trivially prob(R v A) = prob(~R -> A).
So modus ponens does preserve the strength of reason to believe from the premises to the conclusion.
We do have greater reason to believe ~R -> C than we have reason to believe ~R -> A. But that does not contradict that, with either R v A or R v C as the non-logical premise, modus ponens did preserve strength of reason to believe.
Modus ponens is a red herring anyway. It is used by McGee as a needless phony baloney armature for a more simple fact: R v A is equivalent with ~R -> A. That's all we need to know.
So this is the slight of hand that McGee uses to pull off his trick:
(1) He puts the argument into an armature of modus ponens. He makes it seem that the supposedly incorrect inference is the fault of modus ponens. But there is no incorrect inference (strength of reason for belief is preserved from premises to conclusion, trivially as the inference is merely tautological), and the inference doesn't require modus ponens.
(2) He distracts by conflating two different arguments. Yes, ~R - C has greater strength of reason for belief than ~R -> A does, but what is at stake is preservation of strength for belief, not an apples and oranges comparison of the conclusions standing alone.
Here's another interesting example to test your solution on:
Quoting Modus Ponens - Alleged cases of failure - Wikipedia
I think the conclusion is true, and MP is valid here.
For a further twist, consider replacing "Hobbes wrote Hamlet" with "1 = 2".
I agree.
"false -> P" is true.
And again, it's not even about MP:
S v H is equivalent to ~S -> H.
Seems we have agreement that modus ponens is not invalidated by the argument in the OP; that the premises are true, the argument valid and the conclusion true, but incomplete.
'incomplete' is not part of my analysis.
MP is valid.
McGee claims MP is not "strictly valid" which I can only take to mean that MP does not preserve strength of reason to believe. But, contrary to his sleight of hand, his example shows MP preserving strength of reason to believe.
So what does your analysis tell us about Carter?
My analysis doesn't need to say anything about Carter.
All I need to point out is that the conclusion of the example has strength of reason to belief not less than the strength of reason to believe the premises.
But I did mention that the strength of reason to believe ~R -> C is greater than the strength of reason to believe ~R -> A, though that is not needed to refute McGee.
Based on the level of McGee's research in logic and his associations, he must be extremely intelligent and knowledgeable. Not a nano-mote of doubt that I could not possible fathom all the logic he knows. So how could he have made such a rank mistake?
I should have approached the problem more systematically from the start. I thought that the explanation would have to be at a high level involving intensionality and then probability. But then I saw that it is bare bones trivial. That's the magician's trick. He distracts you with a bunch of razzle dazzle hiding the explanation that is the one right in front of your nose.
Then I went on to win the Tour de France, the Indy 500, the Pulitzer Prize, and the Best Pecan Pie Award at the Polk County Fair, all in one week. But everybody already knows all about that, so enough about me.
Quoting TonesInDeepFreeze
But if Reagan did not win, it would have been Carter. And it is this that is in contrast wth the conclusion of the MP. So isn't your explanation incomplete?
Not by the premises of the argument.
The point is not to challenge the premises, but rather to show that the conclusion has as great a reason for belief as the premises. That's all that's needed to refute McGee. And it's trivial. It only looks hard because he razzle dazzles us with a distraction.
I said that I only have the clip from the article to reference plus the locution 'strictly valid'.
Anything I say is in that context alone. If there is more in his article that qualifies the context, then that would be another story.
prob(x) is not presumably the poll rating of x.
For example, if my memory is in the ballpark, the day before the election, Biden was given about an 85% chance of winning but he only had about 52% in the polls.
But my latest posts don't assign prob(x) but only rank the probabilities as given.
prob(R) > prob(C) > prob(A)
And we don't even need that!
Whatever the prob(R v A) is, it's no greater than prob(~R -> A), as indeed prob(R v A) = prob(~R -> A).
And we don't even need that!
All we have to do is see that whatever the strength of reason for believe for R v A, it's no greater than the strength of reason for belief of ~R -> A. How could that not be? R v A is equivalent with ~R -> A. It's that simple!
Directly responding to the clip and its one sentence intro:
MP is valid. McGee does not say MP is not valid. He says it is not "strictly valid". What does "strictly valid" mean? He mentions "good reason to believe". So the only way I can think of regarding "strictly valid" is "If there is good reason to believe the premises then there is good reason to believe the conclusion". He says, of the only non-logical premise, that we have good reason to believe "A Republican will win the election". But "A Republican will win the election" is equivalent to "If it's not Reagan who wins, it will be Anderson". So whatever the good reason we have to believe "A Republican will win the election" is the same good reason we have to believe "If it's not Reagan who wins, it will be Anderson". So this instance of MP that McGee claims shows that strict validity fails, is actually an example in which strict validity succeeds. It's that simple. McGee is refuted.
That's false. But nevermind, we can take it as a hypothetical given. And it doesn't even matter anyway. All we need is the background assumption that people had good reason to believe that a Republican would win. Carter is a red herring.
What people believe in premise and conclusion is irrelevant. People can err by believing the premises of a valid argument but not the conclusion. The strict validity of MP couldn't depend on the empirical fact of what erring humans believe.
So it should be "Those apprised of the poll results had good reason to believe".
[1] is a logical truth and [2] is a non-logical premise.
In the context, [2] is equivalent to "Either Reagan will win or Anderson will win".
If they had good reason to believe [2] then they had good reason to believe [3].
McGee did not show that strict validity of MP failed.
McGee has another supposed impeachment of MP.
https://sites.duke.edu/wsa/papers/files/2011/05/wsa-defenseofmodusponens1986.pdf
"I see what looks like a large fish writhing in a fisherman's net a ways off. I
believe
If that creature is a fish, then if it has lungs, it's a lungfish.
That, after all, is what one means by "lungfish." Yet, even though I
believe the antecedent of this conditional, I do not conclude
If that creature has lungs, it's a lungfish.
Lungfishes are rare, oddly shaped, and, to my knowledge, appear only in
fresh water. It is more likely that, even though it does not look like one,
the animal in the net is a porpoise"
c = that creature
Fx <-> x is a fish
Lx <-> x is a lungfish
Hx <-> x has lungs
(1) Fc -> (Hc -> Lc) [McGee believes]
(2) Fc [McGee believes]
(3) Hc -> Lc [McGee does not believe]
But, again, what people do or do not believe is not relevant. What is relevant is (1) for validity, whether truth is preserved, or I suppose (2) for "strict validity", whether "good reason to believe" is preserved.
Obviously the strength of reason to believe the conclusion Hc -> Lc follows the strength of reason to believe the premise Fc. But we need to show that mathematically. I haven't finished it, but getting close:
prob(Fc -> (Hc -> Lc)) = 100%, since it's true by definition
Let prob(Fc) = x
Let prob(Hc) = y
Let prob(Lc) = z
We also have Lc -> Fc [implicit in the problem; I wish I could dispense with it though]
So prob(Lc) = prob(Fc & Lc) = x*prob(Hc | Fc)
~Hc and Lc are mutually exclusive so:
So prob(Hc -> Lc) = prob(~Hc v Lc) = prob(~Hc)+prob(Lc) = (100-y)+(x*prob(Hc | Fc))
So prob(Hc -> Lc) goes up and down as prob(Fc) goes up and down.
The possibility of the porpoise is not part of this. Just like the possibility of Carter was not part of the earlier problem. The porpoise is Carter! (And, as we know, Paul is the walrus.)
I just now read that. My argument is basically the same as theirs.
Yes, so why do McGee's examples seem to be counterexamples to modus ponens when they are not? Because the way the counterexamples are expressed suggest an unrestricted set of possibilities for the conclusion when, in fact, the possibilities are restricted by the assumptions. D. E. Over explains (note his use of the pronoun 'he'):
Of course the premise "A Republican wins" restricts. The impression that there is not good reason to believe "If Reagan doesn't win then Anderson wins" comes from (1) Overlooking that it "If Reagan doesn't win then Anderson wins" is merely a conditional and not a statement about Anderson winning, and (2) overlooking that "If Reagan doesn't win then Anderson wins" is equivalent with "A Republican wins", so whatever the bases are for believing "A Republican wins" are the same bases for believing "If Reagan doesn't win then Anderson wins".
McGee's error is the claim that the conclusion doesn't have as good a reason to believe as the premises. That is the explicit error in his argument about the MP example.
Indeed Carter is left out with the premise "A Republican wins", but stating that Carter is left out and that therefore the conclusion is clouded is not in and of itself a refutation. The follow-through is that Carter being left out is just "A Republican wins" which is equivalent with the conclusion, so whatever basis there is for "A Republican wins" is bases for the conclusion.
The fact that there is more reason to believe "If Reagan does not win then Carter wins" than there is reason to believe "If Reagan does not win then Carter wins" is McGee's red herring.
There is a difference between (1) What is the best way to set up inferences about the election? and (2) Does the conclusion of the particular MP mentioned have as great a reason for believing as the premises have?
The key to refuting McGee is not (1). It's (2).
McGee also erred by dragging in what people believed. (Probably, nudging in "people believed" is part of the sleight of hand.) What people believed is irrelevant to the example. But we can still couch his argument without concern for what people believed and stick with "reason to believe" only.
/
Another article about the puzzle went into a bunch of stuff about subjunctive mood. For me, that 's a wrong tack: (1) We can rephrase the MP without subjunctive and (2) The puzzle is dissolved much more easily, trivially, anyway.
Yesterday I got hold of McGee's paper.
It turns out that his argument does not suppose that the conditionals mentioned are taken in the sense of the material conditional. He says that if the conditionals mentioned are taken in the sense of the material conditional then modus ponens is not impeached by his argument.
Lack of having his paper to know what he actually claimed led to unnecessary disputation about his argument.
This has been a waste of my time and the time of people reading my posts. If I knew from the onset that he's not talking about the material condition, then I wouldn't have unnecessarily bothered.
Ok, thanks for your work. Glad you found it interesting.
Quoting TonesInDeepFreeze
I don't agree. Solving crossword puzzles is not time wasted.
With good reason, the pollees believe that a Republican will win the election, not that a Republican might win the election, meaning that (1) is invalid.
IE, it is (1) that is not part of a valid modus ponens, rather than the case that the modus ponens is not valid.