# Thread: Predicate logic problem

1. ## Predicate logic problem

∀x:X.¬(P^Q) ¬∃x:X. ¬(¬P v ¬Q)

I want to prove that the left hand side entails () the right hand side using propositional and predicate logic. Thank you

2. ## Re: Predicate logic problem

Do P and Q depend on x? Also, see this sticky thread.

3. ## Re: Predicate logic problem

p and q are of type x and i can use natural deduction and axioms

4. ## Re: Predicate logic problem

Hmm, I am not sure I understand "p and q are of type x." In terms of types, I would say, "P and Q are of type X -> Prop" where Prop is the type of propositions. But I understand that P and Q are unary predicates that accept arguments of type X.

A derivation depends on the axioms you have. For example, if you have an axiom (¬∃x. ¬Ax) <-> ∀x. Ax, then you can immediately use it to derive ¬∃x:X. ¬¬(Px ^ Qx). Similarly, if you have an axiom A ^ B <-> ¬(¬A v ¬B), then you can derive ¬∃x:X. ¬(¬Px v ¬Qx). Otherwise, it is significantly more tricky.

Also, do you use Fitch-style natural deduction (also called flag notation) or tree-like natural deduction?

5. ## Re: Predicate logic problem

i can use the following axioms:

P v ¬P
P ^ ¬P (Proof by contradiction)

So what you re saying is that i can use double negation to remove negation, correct?

6. ## Re: Predicate logic problem

Originally Posted by hmat
P v ¬P
P ^ ¬P (Proof by contradiction)
P ^ ¬P is not an axiom since it is contradictory.

I'll describe the derivation in words. Assume $\exists x\,\neg(\neg Px\lor\neg Qx)$. We need to derive a contradiction. Apply existential elimination to get an x and $\neg(\neg Px\lor\neg Qx)$. Use universal elimination with that x on $\forall x\,\neg(Px\land Qx)$ to get $\neg(Px\land Qx)$. We are going to make two assumptions and show that they are contradictory.

Assume $Px$ and assume $Qx$. Then $Px\land Qx$, which contradicts $\neg(Px\land Qx)$. Therefore, we close $Px$ and derive $\neg Px$. Using disjunction introduction, this implies $\neg Px\lor \neg Qx$, which contradicts $\neg(\neg Px\lor\neg Qx)$. Therefore, we close $Qx$ and derive $\neg Qx$. As before, this implies $\neg Px\lor \neg Qx$, which contradicts $\neg(\neg Px\lor\neg Qx)$. Thus we closed both $Px$ and $Qx$ and derived a contradiction.

Note that I have not used the law of excluded middle $A\lor\neg A$. Using it, it may be possible to make the derivation a little clearer. Basically, once you have $\neg(\neg Px\lor\neg Qx)$ and $\neg(Px\land Qx)$, you consider (using disjunction elimination) four cases where either $Px$ or $\neg Px$ and either $Qx$ or $\neg Qx$ hold. In each case, it is possible to derive a contradiction using the two premises above. This is similar to constructing a truth table.