Theorem ifeq1d

Description: Equality deduction for conditional operator. (Contributed by NM, 16-Feb-2005)

	Ref	Expression
Hypothesis	ifeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	ifeq1d	⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐶 ) = if ( 𝜓 , 𝐵 , 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		ifeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		ifeq1	⊢ ( 𝐴 = 𝐵 → if ( 𝜓 , 𝐴 , 𝐶 ) = if ( 𝜓 , 𝐵 , 𝐶 ) )
3	1 2	syl	⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐶 ) = if ( 𝜓 , 𝐵 , 𝐶 ) )