Metamath Proof Explorer


Theorem ifbieq2d

Description: Equivalence/equality deduction for conditional operators. (Contributed by Paul Chapman, 22-Jun-2011)

Ref Expression
Hypotheses ifbieq2d.1 ( 𝜑 → ( 𝜓𝜒 ) )
ifbieq2d.2 ( 𝜑𝐴 = 𝐵 )
Assertion ifbieq2d ( 𝜑 → if ( 𝜓 , 𝐶 , 𝐴 ) = if ( 𝜒 , 𝐶 , 𝐵 ) )

Proof

Step Hyp Ref Expression
1 ifbieq2d.1 ( 𝜑 → ( 𝜓𝜒 ) )
2 ifbieq2d.2 ( 𝜑𝐴 = 𝐵 )
3 1 ifbid ( 𝜑 → if ( 𝜓 , 𝐶 , 𝐴 ) = if ( 𝜒 , 𝐶 , 𝐴 ) )
4 2 ifeq2d ( 𝜑 → if ( 𝜒 , 𝐶 , 𝐴 ) = if ( 𝜒 , 𝐶 , 𝐵 ) )
5 3 4 eqtrd ( 𝜑 → if ( 𝜓 , 𝐶 , 𝐴 ) = if ( 𝜒 , 𝐶 , 𝐵 ) )