3
$\begingroup$

I have been stuck with this problem for a long time, I tried reductio ad absurdum and I got the hypothesys [¬∃x∀y¬P(x,y)], then I try to eliminate the negation of the premise, but I have to prove ∀x∃yP(x,y), and after using the introduction of universal quantifier rule, I go again with reductio ad absurdum, gaining a second hypothesis [¬∃yP(x,y)]. But at this point I have two hypothesis that contain negations of existencial quantifier, and I don't know how to use them constructively. I found some other similar questions, but all the answers given do not say which rules must be applied, and since I'm a beginner I didn't understand them.

$\endgroup$
0
$\begingroup$

You are doing this exactly right! You just have to derive $\forall y \neg P(x,y)$ from $\neg \exists y P(x,y)$

Now, I am not sure how your proof system defines the rule for $\forall$ Introduction ... in the system that I use you designate a 'fresh' constant to take the role of the arbitrary object from the domain. So this is what it looks like in my preferred system, called Fitch:

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.