Metamath Proof Explorer


Theorem spcegf

Description: Existential specialization, using implicit substitution. (Contributed by NM, 2-Feb-1997)

Ref Expression
Hypotheses spcgf.1 _ x A
spcgf.2 x ψ
spcgf.3 x = A φ ψ
Assertion spcegf A V ψ x φ

Proof

Step Hyp Ref Expression
1 spcgf.1 _ x A
2 spcgf.2 x ψ
3 spcgf.3 x = A φ ψ
4 2 nfn x ¬ ψ
5 3 notbid x = A ¬ φ ¬ ψ
6 1 4 5 spcgf A V x ¬ φ ¬ ψ
7 6 con2d A V ψ ¬ x ¬ φ
8 df-ex x φ ¬ x ¬ φ
9 7 8 syl6ibr A V ψ x φ