Metamath Proof Explorer


Theorem ifeq12d

Description: Equality deduction for conditional operator. (Contributed by NM, 24-Mar-2015)

Ref Expression
Hypotheses ifeq1d.1 ( 𝜑𝐴 = 𝐵 )
ifeq12d.2 ( 𝜑𝐶 = 𝐷 )
Assertion ifeq12d ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐶 ) = if ( 𝜓 , 𝐵 , 𝐷 ) )

Proof

Step Hyp Ref Expression
1 ifeq1d.1 ( 𝜑𝐴 = 𝐵 )
2 ifeq12d.2 ( 𝜑𝐶 = 𝐷 )
3 1 ifeq1d ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐶 ) = if ( 𝜓 , 𝐵 , 𝐶 ) )
4 2 ifeq2d ( 𝜑 → if ( 𝜓 , 𝐵 , 𝐶 ) = if ( 𝜓 , 𝐵 , 𝐷 ) )
5 3 4 eqtrd ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐶 ) = if ( 𝜓 , 𝐵 , 𝐷 ) )