Theorem bicomdd

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)

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

Proof

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