Theorem ifeq2d

Description: Equality deduction for conditional operator. (Contributed by NM, 16-Feb-2005)

	Ref	Expression
Hypothesis	ifeq1d.1	|- ( ph -> A = B )
Assertion	ifeq2d	|- ( ph -> if ( ps , C , A ) = if ( ps , C , B ) )

Proof

Step	Hyp	Ref	Expression
1		ifeq1d.1	|- ( ph -> A = B )
2		ifeq2	|- ( A = B -> if ( ps , C , A ) = if ( ps , C , B ) )
3	1 2	syl	|- ( ph -> if ( ps , C , A ) = if ( ps , C , B ) )