Metamath Proof Explorer


Theorem ifeq1d

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

Ref Expression
Hypothesis ifeq1d.1 φA=B
Assertion ifeq1d φifψAC=ifψBC

Proof

Step Hyp Ref Expression
1 ifeq1d.1 φA=B
2 ifeq1 A=BifψAC=ifψBC
3 1 2 syl φifψAC=ifψBC