Theorem ifbieq1d

Description: Equivalence/equality deduction for conditional operators. (Contributed by JJ, 25-Sep-2018)

	Ref	Expression
Hypotheses	ifbieq1d.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
	ifbieq1d.2	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	ifbieq1d	⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐶 ) = if ( 𝜒 , 𝐵 , 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		ifbieq1d.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
2		ifbieq1d.2	⊢ ( 𝜑 → 𝐴 = 𝐵 )
3	1	ifbid	⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐶 ) = if ( 𝜒 , 𝐴 , 𝐶 ) )
4	2	ifeq1d	⊢ ( 𝜑 → if ( 𝜒 , 𝐴 , 𝐶 ) = if ( 𝜒 , 𝐵 , 𝐶 ) )
5	3 4	eqtrd	⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐶 ) = if ( 𝜒 , 𝐵 , 𝐶 ) )