Metamath Proof Explorer


Theorem ifpnot23

Description: Negation of conditional logical operator. (Contributed by RP, 18-Apr-2020)

Ref Expression
Assertion ifpnot23 ( ¬ if- ( 𝜑 , 𝜓 , 𝜒 ) ↔ if- ( 𝜑 , ¬ 𝜓 , ¬ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 ianor ⊢ ( ¬ ( 𝜑 ∧ 𝜓 ) ↔ ( ¬ 𝜑 ∨ ¬ 𝜓 ) )
2 pm4.55 ⊢ ( ¬ ( ¬ 𝜑 ∧ 𝜒 ) ↔ ( 𝜑 ∨ ¬ 𝜒 ) )
3 1 2 anbi12i ⊢ ( ( ¬ ( 𝜑 ∧ 𝜓 ) ∧ ¬ ( ¬ 𝜑 ∧ 𝜒 ) ) ↔ ( ( ¬ 𝜑 ∨ ¬ 𝜓 ) ∧ ( 𝜑 ∨ ¬ 𝜒 ) ) )
4 ioran ⊢ ( ¬ ( ( 𝜑 ∧ 𝜓 ) ∨ ( ¬ 𝜑 ∧ 𝜒 ) ) ↔ ( ¬ ( 𝜑 ∧ 𝜓 ) ∧ ¬ ( ¬ 𝜑 ∧ 𝜒 ) ) )
5 dfifp4 ⊢ ( if- ( 𝜑 , ¬ 𝜓 , ¬ 𝜒 ) ↔ ( ( ¬ 𝜑 ∨ ¬ 𝜓 ) ∧ ( 𝜑 ∨ ¬ 𝜒 ) ) )
6 3 4 5 3bitr4i ⊢ ( ¬ ( ( 𝜑 ∧ 𝜓 ) ∨ ( ¬ 𝜑 ∧ 𝜒 ) ) ↔ if- ( 𝜑 , ¬ 𝜓 , ¬ 𝜒 ) )
7 df-ifp ⊢ ( if- ( 𝜑 , 𝜓 , 𝜒 ) ↔ ( ( 𝜑 ∧ 𝜓 ) ∨ ( ¬ 𝜑 ∧ 𝜒 ) ) )
8 6 7 xchnxbir ⊢ ( ¬ if- ( 𝜑 , 𝜓 , 𝜒 ) ↔ if- ( 𝜑 , ¬ 𝜓 , ¬ 𝜒 ) )