Theorem ifbid

Description: Equivalence deduction for conditional operators. (Contributed by NM, 18-Apr-2005)

	Ref	Expression
Hypothesis	ifbid.1	|- ( ph -> ( ps <-> ch ) )
Assertion	ifbid	|- ( ph -> if ( ps , A , B ) = if ( ch , A , B ) )

Proof

Step	Hyp	Ref	Expression
1		ifbid.1	|- ( ph -> ( ps <-> ch ) )
2		ifbi	|- ( ( ps <-> ch ) -> if ( ps , A , B ) = if ( ch , A , B ) )
3	1 2	syl	|- ( ph -> if ( ps , A , B ) = if ( ch , A , B ) )