Theorem 3orbi123i 1078

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 668	. . 3 ⊢ ((φ ∨ χ) ↔ (ψ ∨ θ))
4		bi3.3	. . 3 ⊢ (τ ↔ η)
5	3, 4	orbi12i 668	. 2 ⊢ (((φ ∨ χ) ∨ τ) ↔ ((ψ ∨ θ) ∨ η))
6		df-3or 872	. 2 ⊢ ((φ ∨ χ ∨ τ) ↔ ((φ ∨ χ) ∨ τ))
7		df-3or 872	. 2 ⊢ ((ψ ∨ θ ∨ η) ↔ ((ψ ∨ θ) ∨ η))
8	5, 6, 7	3bitr4i 201	1 ⊢ ((φ ∨ χ ∨ τ) ↔ (ψ ∨ θ ∨ η))

Syntax hints:   ↔ wb 98   ∨ wo 616   ∨ w3o 870

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 617

This theorem depends on definitions:  df-bi 110  df-3or 872

This theorem is referenced by: (None)