Theorem or32 687

Description: A rearrangement of disjuncts. (Contributed by NM, 18-Oct-1995.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
or32	⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜓))

Proof of Theorem or32

Step	Hyp	Ref	Expression
1		orass 684	. 2 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
2		or12 683	. 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ (𝜓 ∨ (𝜑 ∨ 𝜒)))
3		orcom 647	. 2 ⊢ ((𝜓 ∨ (𝜑 ∨ 𝜒)) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜓))
4	1, 2, 3	3bitri 195	1 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜓))

Syntax hints:   ↔ wb 98   ∨ wo 629

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630

This theorem depends on definitions:  df-bi 110

This theorem is referenced by:  xrnepnf  8700