Originally Posted by

**logicloser** The problems:

a) $\displaystyle \vdash (\forall x \phi \rightarrow \exists x \phi)$ for any $\displaystyle \phi$

I'm confused about this one because I tried proof by contradiction, so:

(1) $\displaystyle \forall x \phi$ by assumption

(2) $\displaystyle \forall x \neg\phi$ by assuming the negation of $\displaystyle \exists x \phi$ where $\displaystyle \exists x$ is also written as $\displaystyle \neg \forall x \neg$

... and then isn't that already a contradiction?

b) $\displaystyle \forall x(\phi \rightarrow \psi) \vdash (\exists x \phi \rightarrow \psi) $where x is not free in $\displaystyle \psi$

This one I think it must involve axioms 4 and 5 somehow, but I guess that's obvious

c) $\displaystyle \vdash \exists x (x = \tau)$ for any term $\displaystyle \tau$ not involving x

And this one must involve axioms 4 and 6

Derive them using these (for your convenience and my LaTeX practice):

Axioms:

1: $\displaystyle (\phi \rightarrow (\psi \rightarrow \phi))$

2: $\displaystyle ((\phi \rightarrow (\psi \rightarrow \theta)) \rightarrow ((\phi \rightarrow \psi) \rightarrow (\phi \rightarrow \theta)))$

3: $\displaystyle ((\neg\phi\rightarrow\neg\psi)\rightarrow(\psi\rig htarrow\phi))$

4: $\displaystyle (\forall x \phi(x) \rightarrow \phi (\tau))$ ($\displaystyle \tau$ is a term freely substitutable for x in $\displaystyle \phi$)

5: $\displaystyle (\forall x (\phi \rightarrow \psi) \rightarrow (\phi \rightarrow \forall x \psi))$ (where x is not free in $\displaystyle \phi$)

6: $\displaystyle \forall x (x=x)$

7: $\displaystyle (x = y \rightarrow (\phi(x,x) \rightarrow \phi(x,y)))$

As well as Modus Ponens, Generalization, and the Deduction theorem

Any help would be really appreciated!