Theorem orordi 942

Description: Distribution of disjunction over disjunction. (Contributed by NM, 25-Feb-1995.)

Assertion

Ref	Expression
orordi	⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ∨ (𝜑 ∨ 𝜒)))

Proof of Theorem orordi

Step	Hyp	Ref	Expression
1		oridm 918	. . 3 ⊢ ((𝜑 ∨ 𝜑) ↔ 𝜑)
2	1	orbi1i 927	. 2 ⊢ (((𝜑 ∨ 𝜑) ∨ (𝜓 ∨ 𝜒)) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3		or4 940	. 2 ⊢ (((𝜑 ∨ 𝜑) ∨ (𝜓 ∨ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ∨ (𝜑 ∨ 𝜒)))
4	2, 3	bitr3i 280	1 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ∨ (𝜑 ∨ 𝜒)))

This proof depends on syntax axioms:   ↔ wb 209   ∨ wo 861

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-or 862

This theorem is used by:  pm4.78  948  plngrotlem2  29143