Metamath Proof Explorer


Theorem axc5c4c711toc4

Description: Rederivation of axc4 from axc5c4c711 . Note that only propositional calculus is required for the rederivation. (Contributed by Andrew Salmon, 14-Jul-2011) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion axc5c4c711toc4
|- ( A. x ( A. x ph -> ps ) -> ( A. x ph -> A. x ps ) )

Proof

Step Hyp Ref Expression
1 ax-1
 |-  ( A. x ( A. x ph -> ps ) -> ( ph -> A. x ( A. x ph -> ps ) ) )
2 ax-1
 |-  ( ( ph -> A. x ( A. x ph -> ps ) ) -> ( A. x A. x -. A. x A. x ( A. x ph -> ps ) -> ( ph -> A. x ( A. x ph -> ps ) ) ) )
3 axc5c4c711
 |-  ( ( A. x A. x -. A. x A. x ( A. x ph -> ps ) -> ( ph -> A. x ( A. x ph -> ps ) ) ) -> ( A. x ph -> A. x ps ) )
4 1 2 3 3syl
 |-  ( A. x ( A. x ph -> ps ) -> ( A. x ph -> A. x ps ) )