Theorem orddi 1027

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

Assertion

Ref	Expression
orddi	⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ 𝜃)) ↔ (((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)) ∧ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃))))

Proof of Theorem orddi

Step	Hyp	Ref	Expression
1		ordir 1024	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ 𝜃)) ↔ ((𝜑 ∨ (𝜒 ∧ 𝜃)) ∧ (𝜓 ∨ (𝜒 ∧ 𝜃))))
2		ordi 1023	. . 3 ⊢ ((𝜑 ∨ (𝜒 ∧ 𝜃)) ↔ ((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)))
3		ordi 1023	. . 3 ⊢ ((𝜓 ∨ (𝜒 ∧ 𝜃)) ↔ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃)))
4	2, 3	anbi12i 639	. 2 ⊢ (((𝜑 ∨ (𝜒 ∧ 𝜃)) ∧ (𝜓 ∨ (𝜒 ∧ 𝜃))) ↔ (((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)) ∧ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃))))
5	1, 4	bitri 278	1 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ 𝜃)) ↔ (((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)) ∧ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃))))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 400   ∨ wo 860

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-an 401  df-or 861

This theorem is used by:  reuprg  4669  prneimg  4819  wl-cases2-dnf  38195  fzunt  44209  fzuntd  44210  fzunt1d  44211  fzuntgd  44212