Tableau Proof for All x P(x) V Q(x) $\supset$ Exists y P(y) V Exists z Q(z)

[gnome]

To do a tableau proof of this statement:
(\forall x) [P(x) \vee Q(x)] \supset [(\exists x)(P(x) \vee (\exists x)(Q(x)]
I started out by restating it as follows:
(\forall x) [P(x) \vee Q(x)] \supset [(\exists y)(P(y) \vee (\exists z)(Q(z)]
to avoid confusion over what's bound to what (and when).

Is my approach:
valid?
invalid?
recommended?
not?
a good idea?
not?
some other adjective (or expletive)?

 

[Owen Holden]

To do a tableau proof of this statement:
(\forall x) [P(x) \vee Q(x)] \supset [(\exists x)(P(x) \vee (\exists x)(Q(x)]
I started out by restating it as follows:
(\forall x) [P(x) \vee Q(x)] \supset [(\exists y)(P(y) \vee (\exists z)(Q(z)]
to avoid confusion over what's bound to what (and when).

Is my approach:
valid?
invalid?
recommended?
not?
a good idea?
not?
some other adjective (or expletive)?

Your approach is correct.

1. Ey(P(y) v EzQ(z)) <-> (EyP(y) v EzQ(z)),

by, Ey(P(y) v p) <-> (EyP(y) v p).

2. (EyP(y) v EzQ(z)) <-> (ExP(x) v ExQ(x)),

by: EyP(y) <-> ExP(x), EzQ(z) <-> ExQ(x).

3. Ey(P(y) v EzQ(z)) <-> (ExP(x) v ExQ(x)),

by:1, 2, ((p <-> q) & (q <-> r)) -> (p <-> r).

 

[gnome]

Thank you.