Metamath Proof Explorer


Theorem orass

Description: Associative law for disjunction. Theorem *4.33 of WhiteheadRussell p. 118. (Contributed by NM, 5-Aug-1993) (Proof shortened by Andrew Salmon, 26-Jun-2011)

Ref Expression
Assertion orass ( ( ( 𝜑 ∨ 𝜓 ) ∨ 𝜒 ) ↔ ( 𝜑 ∨ ( 𝜓 ∨ 𝜒 ) ) )

Proof

Step Hyp Ref Expression
1 orcom ⊢ ( ( ( 𝜑 ∨ 𝜓 ) ∨ 𝜒 ) ↔ ( 𝜒 ∨ ( 𝜑 ∨ 𝜓 ) ) )
2 or12 ⊢ ( ( 𝜒 ∨ ( 𝜑 ∨ 𝜓 ) ) ↔ ( 𝜑 ∨ ( 𝜒 ∨ 𝜓 ) ) )
3 orcom ⊢ ( ( 𝜒 ∨ 𝜓 ) ↔ ( 𝜓 ∨ 𝜒 ) )
4 3 orbi2i ⊢ ( ( 𝜑 ∨ ( 𝜒 ∨ 𝜓 ) ) ↔ ( 𝜑 ∨ ( 𝜓 ∨ 𝜒 ) ) )
5 1 2 4 3bitri ⊢ ( ( ( 𝜑 ∨ 𝜓 ) ∨ 𝜒 ) ↔ ( 𝜑 ∨ ( 𝜓 ∨ 𝜒 ) ) )