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?