Theorem bicomddOLD 16229

Description: Commute two sides of a biconditional in a deduction.

Hypothesis

Ref	Expression
bicomdd.1	|- (ph -> (ps -> (ch <-> th)))

Assertion

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

Proof of Theorem bicomddOLD

Step	Hyp	Ref	Expression
1		bicomdd.1	. . . 4 |- (ph -> (ps -> (ch <-> th)))
2	1	imp 377	. . 3 |- ((ph /\ ps) -> (ch <-> th))
3	2	bicomd 580	. 2 |- ((ph /\ ps) -> (th <-> ch))
4	3	ex 402	1 |- (ph -> (ps -> (th <-> ch)))

Syntax hints:   -> wi 3   <-> wb 163   /\ wa 240

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7

This theorem depends on definitions:  df-bi 164  df-an 242

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