Theorem bicomdd 34138

Description: Commute two sides of a biconditional in a deduction. (Contributed by Rodolfo Medina, 19-Oct-2010.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)

Hypothesis

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

Assertion

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

Proof of Theorem bicomdd

Step	Hyp	Ref	Expression
1		bicomdd.1	. 2 |- ( ph -> ( ps -> ( ch <-> th ) ) )
2		bicom 212	. 2 |- ( ( ch <-> th ) <-> ( th <-> ch ) )
3	1, 2	syl6ib 241	1 |- ( ph -> ( ps -> ( th <-> ch ) ) )

Syntax hints:   -> wi 4   <-> wb 196

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 197

This theorem is referenced by:  ibdr  34142