Theorem 3orbi123i 1179

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

Hypotheses

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

Assertion

Ref	Expression
3orbi123i	⊢ ((𝜑 ∨ 𝜒 ∨ 𝜏) ↔ (𝜓 ∨ 𝜃 ∨ 𝜂))

Proof of Theorem 3orbi123i

Step	Hyp	Ref	Expression
1		bi3.1	. . . 4 ⊢ (𝜑 ↔ 𝜓)
2		bi3.2	. . . 4 ⊢ (𝜒 ↔ 𝜃)
3	1, 2	orbi12i 754	. . 3 ⊢ ((𝜑 ∨ 𝜒) ↔ (𝜓 ∨ 𝜃))
4		bi3.3	. . 3 ⊢ (𝜏 ↔ 𝜂)
5	3, 4	orbi12i 754	. 2 ⊢ (((𝜑 ∨ 𝜒) ∨ 𝜏) ↔ ((𝜓 ∨ 𝜃) ∨ 𝜂))
6		df-3or 969	. 2 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜏) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜏))
7		df-3or 969	. 2 ⊢ ((𝜓 ∨ 𝜃 ∨ 𝜂) ↔ ((𝜓 ∨ 𝜃) ∨ 𝜂))
8	5, 6, 7	3bitr4i 211	1 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜏) ↔ (𝜓 ∨ 𝜃 ∨ 𝜂))

Syntax hints:   ↔ wb 104   ∨ wo 698   ∨ w3o 967

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699

This theorem depends on definitions:  df-bi 116  df-3or 969

This theorem is referenced by:  nnwetri  6881  exmidontriimlem3  7179