Metamath Proof Explorer


Theorem bj-sblem1

Description: Lemma for substitution. (Contributed by BJ, 23-Jul-2023)

Ref Expression
Assertion bj-sblem1
|- ( A. x ( ph -> ( ps -> ch ) ) -> ( A. x ( ph -> ps ) -> ( E. x ph -> ch ) ) )

Proof

Step Hyp Ref Expression
1 ax-2
 |-  ( ( ph -> ( ps -> ch ) ) -> ( ( ph -> ps ) -> ( ph -> ch ) ) )
2 1 al2imi
 |-  ( A. x ( ph -> ( ps -> ch ) ) -> ( A. x ( ph -> ps ) -> A. x ( ph -> ch ) ) )
3 19.23v
 |-  ( A. x ( ph -> ch ) <-> ( E. x ph -> ch ) )
4 2 3 syl6ib
 |-  ( A. x ( ph -> ( ps -> ch ) ) -> ( A. x ( ph -> ps ) -> ( E. x ph -> ch ) ) )