Metamath Proof Explorer


Theorem ifbieq12d2

Description: Equivalence deduction for conditional operators. (Contributed by Thierry Arnoux, 14-Feb-2017) (Proof shortened by Wolf Lammen, 24-Jun-2021)

Ref Expression
Hypotheses ifbieq12d2.1 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
ifbieq12d2.2 ⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝐴 = 𝐶 )
ifbieq12d2.3 ⊢ ( ( 𝜑 ∧ ¬ 𝜓 ) → 𝐵 = 𝐷 )
Assertion ifbieq12d2 ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐵 ) = if ( 𝜒 , 𝐶 , 𝐷 ) )

Proof

Step Hyp Ref Expression
1 ifbieq12d2.1 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
2 ifbieq12d2.2 ⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝐴 = 𝐶 )
3 ifbieq12d2.3 ⊢ ( ( 𝜑 ∧ ¬ 𝜓 ) → 𝐵 = 𝐷 )
4 2 3 ifeq12da ⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐵 ) = if ( 𝜓 , 𝐶 , 𝐷 ) )
5 1 ifbid ⊢ ( 𝜑 → if ( 𝜓 , 𝐶 , 𝐷 ) = if ( 𝜒 , 𝐶 , 𝐷 ) )
6 4 5 eqtrd ⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐵 ) = if ( 𝜒 , 𝐶 , 𝐷 ) )