Theorem ifbid

Description: Equivalence deduction for conditional operators. (Contributed by NM, 18-Apr-2005)

	Ref	Expression
Hypothesis	ifbid.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion	ifbid	⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐵 ) = if ( 𝜒 , 𝐴 , 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		ifbid.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
2		ifbi	⊢ ( ( 𝜓 ↔ 𝜒 ) → if ( 𝜓 , 𝐴 , 𝐵 ) = if ( 𝜒 , 𝐴 , 𝐵 ) )
3	1 2	syl	⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐵 ) = if ( 𝜒 , 𝐴 , 𝐵 ) )