Metamath Proof Explorer


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