Theorem bisym 281

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 179	. 2 |- ((ch -> th) -> ((th -> ch) -> (ch <-> th)))
2	1	bi3ant 280	1 |- (((ph -> ps) -> (ch -> th)) -> (((ps -> ph) -> (th -> ch)) -> ((ph <-> ps) -> (ch <-> th))))

This proof depends on syntax axioms:   -> wi 4   <-> wb 176

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177

This theorem is used by: (None)