Theorem orimdi 944

Description: Disjunction distributes over implication. (Contributed by Wolf Lammen, 5-Jan-2013.)

Assertion

Ref	Expression
orimdi	⊢ ((𝜑 ∨ (𝜓 → 𝜒)) ↔ ((𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜒)))

Proof of Theorem orimdi

Step	Hyp	Ref	Expression
1		imdi 394	. 2 ⊢ ((¬ 𝜑 → (𝜓 → 𝜒)) ↔ ((¬ 𝜑 → 𝜓) → (¬ 𝜑 → 𝜒)))
2		df-or 862	. 2 ⊢ ((𝜑 ∨ (𝜓 → 𝜒)) ↔ (¬ 𝜑 → (𝜓 → 𝜒)))
3		df-or 862	. . 3 ⊢ ((𝜑 ∨ 𝜓) ↔ (¬ 𝜑 → 𝜓))
4		df-or 862	. . 3 ⊢ ((𝜑 ∨ 𝜒) ↔ (¬ 𝜑 → 𝜒))
5	3, 4	imbi12i 353	. 2 ⊢ (((𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜒)) ↔ ((¬ 𝜑 → 𝜓) → (¬ 𝜑 → 𝜒)))
6	1, 2, 5	3bitr4i 306	1 ⊢ ((𝜑 ∨ (𝜓 → 𝜒)) ↔ ((𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜒)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∨ wo 861

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-or 862

This theorem is used by:  pm2.76  945  pm2.85  946