Theorem pm4.78 948

Description: Implication distributes over disjunction. Theorem *4.78 of [WhiteheadRussell] p. 121. (Contributed by NM, 3-Jan-2005.) (Proof shortened by Wolf Lammen, 19-Nov-2012.)

Assertion

Ref	Expression
pm4.78	⊢ (((𝜑 → 𝜓) ∨ (𝜑 → 𝜒)) ↔ (𝜑 → (𝜓 ∨ 𝜒)))

Proof of Theorem pm4.78

Step	Hyp	Ref	Expression
1		orordi 942	. 2 ⊢ ((¬ 𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ ((¬ 𝜑 ∨ 𝜓) ∨ (¬ 𝜑 ∨ 𝜒)))
2		imor 867	. 2 ⊢ ((𝜑 → (𝜓 ∨ 𝜒)) ↔ (¬ 𝜑 ∨ (𝜓 ∨ 𝜒)))
3		imor 867	. . 3 ⊢ ((𝜑 → 𝜓) ↔ (¬ 𝜑 ∨ 𝜓))
4		imor 867	. . 3 ⊢ ((𝜑 → 𝜒) ↔ (¬ 𝜑 ∨ 𝜒))
5	3, 4	orbi12i 928	. 2 ⊢ (((𝜑 → 𝜓) ∨ (𝜑 → 𝜒)) ↔ ((¬ 𝜑 ∨ 𝜓) ∨ (¬ 𝜑 ∨ 𝜒)))
6	1, 2, 5	3bitr4ri 307	1 ⊢ (((𝜑 → 𝜓) ∨ (𝜑 → 𝜒)) ↔ (𝜑 → (𝜓 ∨ 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: (None)