i'm wondering - how can i use the deduction property to show those two:
1)$\vdash \forall x (p(x) \to q(x)) \to (\exists x p(x) \to \exists x q(x))$
2)$\vdash \forall x (p(x) \to q(x)) \to (\forall x p(x) \to \forall x q(x))$
if there's a better way to prove it (formal proof), i would be interested in learning how to do so.
it seems quite trivial to understand that if $\vdash \forall x (p(x) \to q(x)$ then must exists an x that follows it ($(∃xp(x)→∃xq(x))$ or to put the $\forall$ inside the brackets.
however, i don't know how to prove it and would really appreciate learning how to do so correctly, even though it might seem trivial.
thank you very much for your help
my attempt with 1) after the given answer for 2):
1) $∀x(p(x)→q(x))$ --- assumed
2) $\exists p(x)$ --- assumed
3)$p(c)$
3) $⊢∀x(p(x)→q(x))$ ---
4) $⊢∀x(p(x)→q(x))→(p(c)→q(c))$ --- quantifier axiom, assumption
5) $p(C)→q(C)$ --- assignment
6) q(c) --- from 2) and 5) by Modus Ponens
7) $\exists x p(c)$ ---
8) $\exists x q(c)$ ---
9) $\exists xp(x)→\exists xq(x)$ --- from 2) and 8) by Deduction Theorem
10) $⊢∀x(p(x)→q(x))→(\exists xp(x)→\exists xq(x))$ --- from 1) and 9) by Deduction Theorem
Consider 2) :
1) $∀x(p(x) → q(x))$ --- assumed
2) $∀xp(x)$ --- assumed
3) $\vdash ∀x(p(x) → q(x)) \to (p(x) → q(x))$ --- quantifier axiom
4) $p(x) → q(x)$ --- from 1) and 3) by Modus Ponens
5) $\vdash ∀xp(x) \to p(x)$ --- quantifier axiom
6) $p(x)$ --- from 2) and 5) by Modus Ponens
7) $q(x)$ --- from 4) and 6) by Modus Ponens
8) $∀xq(x)$ --- from 7) by Generalization
9) $∀xp(x) \to ∀xq(x)$ --- from 2) and 8) by Deduction Theorem
For axioms and rules, see E.Mendelson, Introduction to Mathematical Logic, Ch,2 First-Order Logic.
We can prove 1) with Hilbert-calculus using the equivalence :
From $∀x(p(x)→q(x))$ we derive $p(x)→q(x)$ and then $¬q(x)→¬p(x)$.
Then we have to use Generalization and then apply 2) to get $∀x¬q(x) → ∀x¬p(x)$.
Finally, we use again contraposition to get :