Metamath Proof Explorer


Theorem 3orbi123i

Description: Join 3 biconditionals with disjunction. (Contributed by NM, 17-May-1994)

Ref Expression
Hypotheses bi3.1 ⊢ ( 𝜑 ↔ 𝜓 )
bi3.2 ⊢ ( 𝜒 ↔ 𝜃 )
bi3.3 ⊢ ( 𝜏 ↔ 𝜂 )
Assertion 3orbi123i ( ( 𝜑 ∨ 𝜒 ∨ 𝜏 ) ↔ ( 𝜓 ∨ 𝜃 ∨ 𝜂 ) )

Proof

Step Hyp Ref Expression
1 bi3.1 ⊢ ( 𝜑 ↔ 𝜓 )
2 bi3.2 ⊢ ( 𝜒 ↔ 𝜃 )
3 bi3.3 ⊢ ( 𝜏 ↔ 𝜂 )
4 1 2 orbi12i ⊢ ( ( 𝜑 ∨ 𝜒 ) ↔ ( 𝜓 ∨ 𝜃 ) )
5 4 3 orbi12i ⊢ ( ( ( 𝜑 ∨ 𝜒 ) ∨ 𝜏 ) ↔ ( ( 𝜓 ∨ 𝜃 ) ∨ 𝜂 ) )
6 df-3or ⊢ ( ( 𝜑 ∨ 𝜒 ∨ 𝜏 ) ↔ ( ( 𝜑 ∨ 𝜒 ) ∨ 𝜏 ) )
7 df-3or ⊢ ( ( 𝜓 ∨ 𝜃 ∨ 𝜂 ) ↔ ( ( 𝜓 ∨ 𝜃 ) ∨ 𝜂 ) )
8 5 6 7 3bitr4i ⊢ ( ( 𝜑 ∨ 𝜒 ∨ 𝜏 ) ↔ ( 𝜓 ∨ 𝜃 ∨ 𝜂 ) )