Theorem biimpor 38066

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

Assertion

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

Proof of Theorem biimpor

Step	Hyp	Ref	Expression
1		imor 853	. 2 ⊢ (((𝜑 ↔ 𝜓) → 𝜒) ↔ (¬ (𝜑 ↔ 𝜓) ∨ 𝜒))
2		notbinot2 38065	. . 3 ⊢ (¬ (𝜑 ↔ 𝜓) ↔ (¬ 𝜑 ↔ 𝜓))
3	2	orbi1i 913	. 2 ⊢ ((¬ (𝜑 ↔ 𝜓) ∨ 𝜒) ↔ ((¬ 𝜑 ↔ 𝜓) ∨ 𝜒))
4	1, 3	bitri 275	1 ⊢ (((𝜑 ↔ 𝜓) → 𝜒) ↔ ((¬ 𝜑 ↔ 𝜓) ∨ 𝜒))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∨ wo 847

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 207  df-or 848

This theorem is referenced by: (None)