Theorem bisym 214

Description: Express symmetries of theorems in terms of biconditionals. (Contributed by Wolf Lammen, 14-May-2013.)

Assertion

Ref	Expression
bisym	|- ( ( ( ph -> ps ) -> ( ch -> th ) ) -> ( ( ( ps -> ph ) -> ( th -> ch ) ) -> ( ( ph <-> ps ) -> ( ch <-> th ) ) ) )

Proof of Theorem bisym

Step	Hyp	Ref	Expression
1		bi3 112	. 2 |- ( ( ch -> th ) -> ( ( th -> ch ) -> ( ch <-> th ) ) )
2	1	bi3ant 213	1 |- ( ( ( ph -> ps ) -> ( ch -> th ) ) -> ( ( ( ps -> ph ) -> ( th -> ch ) ) -> ( ( ph <-> ps ) -> ( ch <-> th ) ) ) )

Syntax hints:   -> wi 4   <-> wb 98

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101

This theorem depends on definitions:  df-bi 110

This theorem is referenced by: (None)