Theorem ordi 1002

Description: Distributive law for disjunction. Theorem *4.41 of [WhiteheadRussell] p. 119. (Contributed by NM, 5-Jan-1993.) (Proof shortened by Andrew Salmon, 7-May-2011.) (Proof shortened by Wolf Lammen, 28-Nov-2013.)

Assertion

Ref	Expression
ordi	⊢ ((𝜑 ∨ (𝜓 ∧ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ∧ (𝜑 ∨ 𝜒)))

Proof of Theorem ordi

Step	Hyp	Ref	Expression
1		jcab 520	. 2 ⊢ ((¬ 𝜑 → (𝜓 ∧ 𝜒)) ↔ ((¬ 𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))
2		df-or 844	. 2 ⊢ ((𝜑 ∨ (𝜓 ∧ 𝜒)) ↔ (¬ 𝜑 → (𝜓 ∧ 𝜒)))
3		df-or 844	. . 3 ⊢ ((𝜑 ∨ 𝜓) ↔ (¬ 𝜑 → 𝜓))
4		df-or 844	. . 3 ⊢ ((𝜑 ∨ 𝜒) ↔ (¬ 𝜑 → 𝜒))
5	3, 4	anbi12i 628	. 2 ⊢ (((𝜑 ∨ 𝜓) ∧ (𝜑 ∨ 𝜒)) ↔ ((¬ 𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))
6	1, 2, 5	3bitr4i 305	1 ⊢ ((𝜑 ∨ (𝜓 ∧ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ∧ (𝜑 ∨ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398   ∨ wo 843

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 209  df-an 399  df-or 844

This theorem is referenced by:  ordir  1003  orddi  1006  pm5.63  1016  pm4.43  1019  cadan  1606  undi  4250  undif3  4264  undif4  4415  elnn1uz2  12324  or3di  30224  ifpan23  39823  ifpidg  39855  ifpim123g  39864