Theorem or3dir 28692

Description: Distributive law for disjunction. (Contributed by Thierry Arnoux, 3-Jul-2017.)

Assertion

Ref	Expression
or3dir	⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∨ 𝜏) ↔ ((𝜑 ∨ 𝜏) ∧ (𝜓 ∨ 𝜏) ∧ (𝜒 ∨ 𝜏)))

Proof of Theorem or3dir

Step	Hyp	Ref	Expression
1		or3di 28691	. 2 ⊢ ((𝜏 ∨ (𝜑 ∧ 𝜓 ∧ 𝜒)) ↔ ((𝜏 ∨ 𝜑) ∧ (𝜏 ∨ 𝜓) ∧ (𝜏 ∨ 𝜒)))
2		orcom 401	. 2 ⊢ ((𝜏 ∨ (𝜑 ∧ 𝜓 ∧ 𝜒)) ↔ ((𝜑 ∧ 𝜓 ∧ 𝜒) ∨ 𝜏))
3		orcom 401	. . 3 ⊢ ((𝜏 ∨ 𝜑) ↔ (𝜑 ∨ 𝜏))
4		orcom 401	. . 3 ⊢ ((𝜏 ∨ 𝜓) ↔ (𝜓 ∨ 𝜏))
5		orcom 401	. . 3 ⊢ ((𝜏 ∨ 𝜒) ↔ (𝜒 ∨ 𝜏))
6	3, 4, 5	3anbi123i 1244	. 2 ⊢ (((𝜏 ∨ 𝜑) ∧ (𝜏 ∨ 𝜓) ∧ (𝜏 ∨ 𝜒)) ↔ ((𝜑 ∨ 𝜏) ∧ (𝜓 ∨ 𝜏) ∧ (𝜒 ∨ 𝜏)))
7	1, 2, 6	3bitr3i 289	1 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∨ 𝜏) ↔ ((𝜑 ∨ 𝜏) ∧ (𝜓 ∨ 𝜏) ∧ (𝜒 ∨ 𝜏)))

Syntax hints:   ↔ wb 195   ∨ wo 382   ∧ w3a 1031

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 196  df-or 384  df-an 385  df-3an 1033

This theorem is referenced by: (None)