Theorem ifeq1d

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

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

Proof

Step	Hyp	Ref	Expression
1		ifeq1d.1	|- ( ph -> A = B )
2		ifeq1	|- ( A = B -> if ( ps , A , C ) = if ( ps , B , C ) )
3	1 2	syl	|- ( ph -> if ( ps , A , C ) = if ( ps , B , C ) )