Theorem 3orbi123d 1251

Description: Deduction joining 3 equivalences to form equivalence of disjunctions. (Contributed by NM, 20-Apr-1994.)

Hypotheses

Ref	Expression
bi3d.1	⊢ (φ → (ψ ↔ χ))
bi3d.2	⊢ (φ → (θ ↔ τ))
bi3d.3	⊢ (φ → (η ↔ ζ))

Assertion

Ref	Expression
3orbi123d	⊢ (φ → ((ψ ∨ θ ∨ η) ↔ (χ ∨ τ ∨ ζ)))

Proof of Theorem 3orbi123d

Step	Hyp	Ref	Expression
1		bi3d.1	. . . 4 ⊢ (φ → (ψ ↔ χ))
2		bi3d.2	. . . 4 ⊢ (φ → (θ ↔ τ))
3	1, 2	orbi12d 690	. . 3 ⊢ (φ → ((ψ ∨ θ) ↔ (χ ∨ τ)))
4		bi3d.3	. . 3 ⊢ (φ → (η ↔ ζ))
5	3, 4	orbi12d 690	. 2 ⊢ (φ → (((ψ ∨ θ) ∨ η) ↔ ((χ ∨ τ) ∨ ζ)))
6		df-3or 935	. 2 ⊢ ((ψ ∨ θ ∨ η) ↔ ((ψ ∨ θ) ∨ η))
7		df-3or 935	. 2 ⊢ ((χ ∨ τ ∨ ζ) ↔ ((χ ∨ τ) ∨ ζ))
8	5, 6, 7	3bitr4g 279	1 ⊢ (φ → ((ψ ∨ θ ∨ η) ↔ (χ ∨ τ ∨ ζ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∨ wo 357   ∨ w3o 933

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-or 359  df-3or 935

This theorem is used by:  moeq3  3014  ltfintri  4467  nncdiv3  6278