Given:
1) [math]\forall x[x=x][/math]2)[math][P(0)\wedge\forall x(P(x)\implies P(x'))]\implies\forall x(P(x))[/math]Then prove in the predicate calculus:
[math]\forall x[(x=0\vee\exists y( x=y')][/math]
Here the P((x) is [math][(x=0\vee\exists y( x=y')][/math]
You can consider the above as a challenge question
1) [math]\forall x[x=x][/math]2)[math][P(0)\wedge\forall x(P(x)\implies P(x'))]\implies\forall x(P(x))[/math]Then prove in the predicate calculus:
[math]\forall x[(x=0\vee\exists y( x=y')][/math]
Here the P((x) is [math][(x=0\vee\exists y( x=y')][/math]
You can consider the above as a challenge question