Metamath Proof Explorer


Theorem ifbid

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

Ref Expression
Hypothesis ifbid.1 φψχ
Assertion ifbid φifψAB=ifχAB

Proof

Step Hyp Ref Expression
1 ifbid.1 φψχ
2 ifbi ψχifψAB=ifχAB
3 1 2 syl φifψAB=ifχAB