1. ## Formal Logic Proof

Hello.

I posted this on math help boards:

(http:///f15/formal-logic-proof-3601/#post15926)

...but I didn't know how active that site was. Feel free to reply on there as well as here.

This is my question:

Give a formal proof to show $\displaystyle \forall x (0' + x' ) = (x . 0'') \vdash \exists x (x + x')= (x . x')$

I'm new to these, and this one looks like it should be easy.

What I want to do is:
1). substitute x into where there are already x's.
2). Make the statement valid for all y
3). substitute y into 0' and y' into 0''.
3). Since it's valid for all y, choose y to be 0.

Here's what I did:

1 (1) $\displaystyle \forall x (0' + x' ) = (x . 0'')$ Assumption
1 (2) $\displaystyle (0' +x')=(x.0'')$ Universal Elimination rule
1 (3) $\displaystyle \exists y (y+x')=(x.y')$ Existential Introduction, 2
4 (4) $\displaystyle y=x$ Assumption
1,4 (5) $\displaystyle \exists x (x+x')=(x.x')$ Taut 3,4 <--- ?
1 (6) $\displaystyle \exists x (x+x')= (x.x')$ Existential Hypothesis, 5

I think i've got the right idea, I think the execution starts to go wrong at around line (4).

Does anyone have any ideas?