Theorem ordir 1001

Description: Distributive law for disjunction. (Contributed by NM, 12-Aug-1994.)

Assertion

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

Proof of Theorem ordir

Step	Hyp	Ref	Expression
1		ordi 1000	. 2 ⊢ ((𝜒 ∨ (𝜑 ∧ 𝜓)) ↔ ((𝜒 ∨ 𝜑) ∧ (𝜒 ∨ 𝜓)))
2		orcom 865	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ 𝜒) ↔ (𝜒 ∨ (𝜑 ∧ 𝜓)))
3		orcom 865	. . 3 ⊢ ((𝜑 ∨ 𝜒) ↔ (𝜒 ∨ 𝜑))
4		orcom 865	. . 3 ⊢ ((𝜓 ∨ 𝜒) ↔ (𝜒 ∨ 𝜓))
5	3, 4	anbi12i 626	. 2 ⊢ (((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)) ↔ ((𝜒 ∨ 𝜑) ∧ (𝜒 ∨ 𝜓)))
6	1, 2, 5	3bitr4i 304	1 ⊢ (((𝜑 ∧ 𝜓) ∨ 𝜒) ↔ ((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)))

Syntax hints:   ↔ wb 207   ∧ wa 396   ∨ wo 842

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 208  df-an 397  df-or 843

This theorem is referenced by:  orddi  1004  pm5.62  1013  dn1  1050  cadan  1595  elnn0z  11848  ifpim123g  39372  rp-fakeanorass  39385  fvmptrabdm  43030