predicate calculus

[chrislav]

Given:
1) $$\forall x[x=x]$$2)$$[P(0)\wedge\forall x(P(x)\implies P(x'))]\implies\forall x(P(x))$$Then prove in the predicate calculus:

$$\forall x[(x=0\vee\exists y( x=y')]$$
Here the P((x) is $$[(x=0\vee\exists y( x=y')]$$
You can consider the above as a challenge question