Metamath Proof Explorer


Theorem ifeq2d

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

Ref Expression
Hypothesis ifeq1d.1 ( 𝜑𝐴 = 𝐵 )
Assertion ifeq2d ( 𝜑 → if ( 𝜓 , 𝐶 , 𝐴 ) = if ( 𝜓 , 𝐶 , 𝐵 ) )

Proof

Step Hyp Ref Expression
1 ifeq1d.1 ( 𝜑𝐴 = 𝐵 )
2 ifeq2 ( 𝐴 = 𝐵 → if ( 𝜓 , 𝐶 , 𝐴 ) = if ( 𝜓 , 𝐶 , 𝐵 ) )
3 1 2 syl ( 𝜑 → if ( 𝜓 , 𝐶 , 𝐴 ) = if ( 𝜓 , 𝐶 , 𝐵 ) )