Metamath Proof Explorer


Theorem biimpor

Description: A rewriting rule for biconditional. (Contributed by Giovanni Mascellani, 15-Sep-2017)

Ref Expression
Assertion biimpor ( ( ( 𝜑 ↔ 𝜓 ) → 𝜒 ) ↔ ( ( ¬ 𝜑 ↔ 𝜓 ) ∨ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 imor ⊢ ( ( ( 𝜑 ↔ 𝜓 ) → 𝜒 ) ↔ ( ¬ ( 𝜑 ↔ 𝜓 ) ∨ 𝜒 ) )
2 notbinot2 ⊢ ( ¬ ( 𝜑 ↔ 𝜓 ) ↔ ( ¬ 𝜑 ↔ 𝜓 ) )
3 2 orbi1i ⊢ ( ( ¬ ( 𝜑 ↔ 𝜓 ) ∨ 𝜒 ) ↔ ( ( ¬ 𝜑 ↔ 𝜓 ) ∨ 𝜒 ) )
4 1 3 bitri ⊢ ( ( ( 𝜑 ↔ 𝜓 ) → 𝜒 ) ↔ ( ( ¬ 𝜑 ↔ 𝜓 ) ∨ 𝜒 ) )