Exercise from Bonevac's "Deduction"

Jose Guilherme February 21, 2019 at 18:02 2950 views 6 comments

Hello

I’m studying by myself Daniel Bonevac’s book “Deduction” and I’ve encountered some exercises whose resolution I’m finding problematic. There are some that I would like to discuss with you, but for now I would like to concentrate on exercise 20, page 238.

So, the exercise is the following:

3xVy( y = x <-> Gy); therefore 3x3yVz ( ~ (x = y) & (z = x V z = y)) <-> 3xVy( y = x <-> ~ Gy)

Any comment would be greatly appreciated.

Comments (6)

Deleted User February 21, 2019 at 18:40 #258196 0 likes

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Jose Guilherme February 21, 2019 at 18:46 #258197 0 likes

Oh... my bad (I'm portuguese, by the way, so some errors in my english are possible).

<-> is the conective for the logical equivalence. So in the exercise "y = x <-> Gy" is "y is equal to x if and ony if y is G".

fdrake February 21, 2019 at 20:21 #258219 0 likes

3xVy( y = x <-> Gy); therefore 3x3yVz ( ~ (x = y) & (z = x V z = y)) <-> 3xVy( y = x <-> ~ Gy)

=

[math]\exists x \forall y (y=x \leftrightarrow Gy) \vdash \exists x \exists y \forall z [((\neg(x=y) \wedge (z=x \vee z=y))]\leftrightarrow \exists x \forall y (y=x \leftrightarrow \neg Gy) [/math]

?

Forum has mathjax support.

Jose Guilherme February 21, 2019 at 20:53 #258226 0 likes

Yes.

It's precisely that!

andrewk February 21, 2019 at 21:44 #258242 0 likes

Let's try making it a bit more intuitive.

Let the set of all objects be U. We don't know what properties the predicate G has, but we know that, like any unary predicate, it must divide U into a set of objects that satisfy G, call it u1, and a set that doesn't, call it u2.

Then the first statement, call it P1 tells us that only one object satisfies G, ie that u1 consists of exactly one object.

The statement we are asked to prove, call it P2, has two parts, call them A and B, that we are asked to prove are logically equivalent, ie that the <-> holds, given the assumption P1.

Inspecting A carefully, we see that it says that U consists of exactly two objects.
Inspecting B carefully, we see that it says that u2 consists of exactly one object.

Stepping back and thinking of a universe containing only two objects, it seems plausible that P1 would imply that A and B are equivalent. Now we just need to prove it formally.

As is usual with proofs of equivalence (<->), we chould first prove one direction, then the other. Choose the easiest-looking direction first.

Jose Guilherme February 22, 2019 at 11:18 #258362 0 likes

So I will show you where currently lies my problem. So, I’m trying a reduction but I’m not seeing how to get the contradiction. I’m guessing that I’ve to get a proposition telling “ ~ (b =a)”, but I’m not seeing how.

https://www.dropbox.com/s/nw0cmgx0bqvybml/IMG_0059.JPG?dl=0