Second-order logic question
Second-order logic is very popular for professional philosophers and logicians. However, it is not much discussed in logic intro textbooks. I want to know how to express 'there is no property F' in second-order logic. I tried something:
~(?X)(?x)(Xx ? Fx)
~(?X)(?x)(Xx ? Fx)
I am not sure which one is correct. Maybe both are incorrect. Can anyone help?
~(?X)(?x)(Xx ? Fx)
~(?X)(?x)(Xx ? Fx)
I am not sure which one is correct. Maybe both are incorrect. Can anyone help?
Comments (2)
Same holds for the second.
It's tricky what you mean by a property not existing: you can of course say that no individual bears the property, but that can be done in first-order. The fact that you are using the symbol F at all already commits your ontology to the property denoted by F to 'existing' in the minimal sense that it's in the domain of quantification of property-variables. If you wanted to say it didn't exist while quantifying over it, you'd have to have some alternate notion of existence, maybe expressed by an existence predicate on properties (that might return true just in case the property holds of no individual, or cannot hold of any individual).
It's the same puzzle about how to express that a certain individual doesn't exist in first-order logic. Suppose you write something like:
~(?x)(x = a)
But that requires the individual denoted by 'a' to be in your domain, and so in the domain of quantification of individual variables. Even to say that whatever 'a' denotes doesn't exist, then, is to commit yourself to its 'existence' in this minimal (Meinongian) sense.
Thanks for the answer.