Metamath Proof Explorer


Theorem ifeq1d

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

Ref Expression
Hypothesis ifeq1d.1 φ A = B
Assertion ifeq1d φ if ψ A C = if ψ B C

Proof

Step Hyp Ref Expression
1 ifeq1d.1 φ A = B
2 ifeq1 A = B if ψ A C = if ψ B C
3 1 2 syl φ if ψ A C = if ψ B C