Metamath Proof Explorer


Theorem spcimgf

Description: Rule of specialization, using implicit substitution. Compare Theorem 7.3 of Quine p. 44. (Contributed by Mario Carneiro, 4-Jan-2017)

Ref Expression
Hypotheses spcimgf.1
|- F/_ x A
spcimgf.2
|- F/ x ps
spcimgf.3
|- ( x = A -> ( ph -> ps ) )
Assertion spcimgf
|- ( A e. V -> ( A. x ph -> ps ) )

Proof

Step Hyp Ref Expression
1 spcimgf.1
 |-  F/_ x A
2 spcimgf.2
 |-  F/ x ps
3 spcimgf.3
 |-  ( x = A -> ( ph -> ps ) )
4 2 1 spcimgft
 |-  ( A. x ( x = A -> ( ph -> ps ) ) -> ( A e. V -> ( A. x ph -> ps ) ) )
5 4 3 mpg
 |-  ( A e. V -> ( A. x ph -> ps ) )