Metamath Proof Explorer


Theorem spcimegf

Description: Existential specialization, using implicit substitution. (Contributed by Mario Carneiro, 4-Jan-2017)

Ref Expression
Hypotheses spcimgf.1 _ x A
spcimgf.2 x ψ
spcimegf.3 x = A ψ φ
Assertion spcimegf A V ψ x φ

Proof

Step Hyp Ref Expression
1 spcimgf.1 _ x A
2 spcimgf.2 x ψ
3 spcimegf.3 x = A ψ φ
4 2 nfn x ¬ ψ
5 3 con3d x = A ¬ φ ¬ ψ
6 1 4 5 spcimgf 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