Metamath Proof Explorer


Theorem ifpfal

Description: Value of the conditional operator for propositions when its first argument is false. Analogue for propositions of iffalse . This is essentially dedlemb . (Contributed by BJ, 20-Sep-2019) (Proof shortened by Wolf Lammen, 25-Jun-2020)

Ref Expression
Assertion ifpfal ( ¬ 𝜑 → ( if- ( 𝜑 , 𝜓 , 𝜒 ) ↔ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 ifpn ( if- ( 𝜑 , 𝜓 , 𝜒 ) ↔ if- ( ¬ 𝜑 , 𝜒 , 𝜓 ) )
2 ifptru ( ¬ 𝜑 → ( if- ( ¬ 𝜑 , 𝜒 , 𝜓 ) ↔ 𝜒 ) )
3 1 2 syl5bb ( ¬ 𝜑 → ( if- ( 𝜑 , 𝜓 , 𝜒 ) ↔ 𝜒 ) )