Theorem ordi 990

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 507	. 2 ⊢ ((¬ 𝜑 → (𝜓 ∧ 𝜒)) ↔ ((¬ 𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))
2		df-or 837	. 2 ⊢ ((𝜑 ∨ (𝜓 ∧ 𝜒)) ↔ (¬ 𝜑 → (𝜓 ∧ 𝜒)))
3		df-or 837	. . 3 ⊢ ((𝜑 ∨ 𝜓) ↔ (¬ 𝜑 → 𝜓))
4		df-or 837	. . 3 ⊢ ((𝜑 ∨ 𝜒) ↔ (¬ 𝜑 → 𝜒))
5	3, 4	anbi12i 612	. 2 ⊢ (((𝜑 ∨ 𝜓) ∧ (𝜑 ∨ 𝜒)) ↔ ((¬ 𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))
6	1, 2, 5	3bitr4i 292	1 ⊢ ((𝜑 ∨ (𝜓 ∧ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ∧ (𝜑 ∨ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 196   ∧ wa 382   ∨ wo 836

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 197  df-an 383  df-or 837

This theorem is referenced by:  ordir  991  orddi  994  pm5.63  1005  pm4.43  1008  cadan  1696  undi  4023  undif3  4037  undif4  4178  elnn1uz2  11973  or3di  29647  ifpan23  38330  ifpidg  38362  ifpim123g  38371