Theorem ifpn

Description: Conditional operator for the negation of a proposition. (Contributed by BJ, 30-Sep-2019) (Proof shortened by Wolf Lammen, 5-May-2024)

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

Proof

Step	Hyp	Ref	Expression
1		ancom	⊢ ( ( ( ¬ 𝜑 ∨ 𝜓 ) ∧ ( ¬ 𝜑 → 𝜒 ) ) ↔ ( ( ¬ 𝜑 → 𝜒 ) ∧ ( ¬ 𝜑 ∨ 𝜓 ) ) )
2		dfifp5	⊢ ( if- ( 𝜑 , 𝜓 , 𝜒 ) ↔ ( ( ¬ 𝜑 ∨ 𝜓 ) ∧ ( ¬ 𝜑 → 𝜒 ) ) )
3		dfifp3	⊢ ( if- ( ¬ 𝜑 , 𝜒 , 𝜓 ) ↔ ( ( ¬ 𝜑 → 𝜒 ) ∧ ( ¬ 𝜑 ∨ 𝜓 ) ) )
4	1 2 3	3bitr4i	⊢ ( if- ( 𝜑 , 𝜓 , 𝜒 ) ↔ if- ( ¬ 𝜑 , 𝜒 , 𝜓 ) )