Theorem 3orbi123i 1220

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 776	. . 3 ⊢ ((𝜑 ∨ 𝜒) ↔ (𝜓 ∨ 𝜃))
4		bi3.3	. . 3 ⊢ (𝜏 ↔ 𝜂)
5	3, 4	orbi12i 776	. 2 ⊢ (((𝜑 ∨ 𝜒) ∨ 𝜏) ↔ ((𝜓 ∨ 𝜃) ∨ 𝜂))
6		df-3or 1010	. 2 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜏) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜏))
7		df-3or 1010	. 2 ⊢ ((𝜓 ∨ 𝜃 ∨ 𝜂) ↔ ((𝜓 ∨ 𝜃) ∨ 𝜂))
8	5, 6, 7	3bitr4i 212	1 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜏) ↔ (𝜓 ∨ 𝜃 ∨ 𝜂))

This proof depends on syntax axioms:   ↔ wb 105   ∨ wo 720   ∨ w3o 1008

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721

This proof depends on definitions:  df-bi 117  df-3or 1010

This theorem is used by:  nnwetri  7223  exmidontriimlem3  7580