# Thread: Symbolic Logic proof problem

1. ## Symbolic Logic proof problem

Hello, I am having a good deal of trouble solving this using the natural deduction rules. If someone could help me by showing the steps involved, that would be great. The argument is as follows: (x)(Fx ⊃(Gx • (y)Hy)) (Therefore) (∃y)(x)(Fx ⊃(Gx •Hy)) The nested y variable is confusing me, and I am not really sure how to go about solving this one using UI, EG, etc.
Btw, the methods that need to be used are those found in Copi's book, using the rules of inference, replacement, and quantification.

2. ## Re: Symbolic Logic proof problem

I'm not totally familiar with the notation you are using. Does the bullet • mean AND (conjunction) or OR (disjunction)? I believe the argument is invalid whether or not your • means AND or OR.

This is what I think the argument is:

Premise: For all x, if Fx, then it is the case that Gx OR/AND for all y, Hy.
Conclusion: There exists y such that for all x, if Fx, then it is the case that Gx OR/AND Hy.

In order to show that the argument is not valid, we have to find one instance in which the premise is true but the conclusion is false. If we take the statement "for all x, if Fx" to be true and the statement "for all y, Hy" to be false, then we can see that the conclusion is obviously false whereas the premise is vacuously true. This shows that the argument is untenable.

3. ## Re: Symbolic Logic proof problem

Originally Posted by Sylvia104
I'm not totally familiar with the notation you are using. Does the bullet • mean AND (conjunction) or OR (disjunction)?
The bullet (unfortunately) means conjunction in "Introduction to Logic" by I. Copi and C. Cohen.

Originally Posted by Sylvia104
If we take the statement "for all x, if Fx" to be true and the statement "for all y, Hy" to be false, then we can see that the conclusion is obviously false whereas the premise is vacuously true.
The premise is also false in this case. The original argument is valid, but only in nonempty models because it involves showing that (y)Hy ⊃ (∃y)Hy.