Theorem 3orbi123i 1245

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 542	. . 3 ⊢ ((𝜑 ∨ 𝜒) ↔ (𝜓 ∨ 𝜃))
4		bi3.3	. . 3 ⊢ (𝜏 ↔ 𝜂)
5	3, 4	orbi12i 542	. 2 ⊢ (((𝜑 ∨ 𝜒) ∨ 𝜏) ↔ ((𝜓 ∨ 𝜃) ∨ 𝜂))
6		df-3or 1032	. 2 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜏) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜏))
7		df-3or 1032	. 2 ⊢ ((𝜓 ∨ 𝜃 ∨ 𝜂) ↔ ((𝜓 ∨ 𝜃) ∨ 𝜂))
8	5, 6, 7	3bitr4i 291	1 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜏) ↔ (𝜓 ∨ 𝜃 ∨ 𝜂))

Syntax hints:   ↔ wb 195   ∨ wo 382   ∨ w3o 1030

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 196  df-or 384  df-3or 1032

This theorem is referenced by:  ne3anior  2875  wecmpep  5030  cnvso  5591  sorpss  6840  ordon  6874  soxp  7177  dford2  8400  axlowdimlem6  25627  elxrge02  28971  brtp  30892  socnv  30908  dfon2  30941  sltsolem1  31067  frege129d  37074  elfz0lmr  40367