Theorem bi3ord 635

Description: Deduction joining 3 equivalences to form equivalence of disjunctions.

Hypotheses

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

Assertion

Ref	Expression
bi3ord	⊢ (φ → ((ψ ∨ θ ∨ η) ↔ (χ ∨ τ ∨ ζ)))

Proof of Theorem bi3ord

Step	Hyp	Ref	Expression
1		bi3d.1	. . . 4 ⊢ (φ → (ψ ↔ χ))
2		bi3d.2	. . . 4 ⊢ (φ → (θ ↔ τ))
3	1, 2	orbi12d 475	. . 3 ⊢ (φ → ((ψ ∨ θ) ↔ (χ ∨ τ)))
4		bi3d.3	. . 3 ⊢ (φ → (η ↔ ζ))
5	3, 4	orbi12d 475	. 2 ⊢ (φ → (((ψ ∨ θ) ∨ η) ↔ ((χ ∨ τ) ∨ ζ)))
6		df-3or 582	. 2 ⊢ ((ψ ∨ θ ∨ η) ↔ ((ψ ∨ θ) ∨ η))
7		df-3or 582	. 2 ⊢ ((χ ∨ τ ∨ ζ) ↔ ((χ ∨ τ) ∨ ζ))
8	5, 6, 7	3bitr4g 428	1 ⊢ (φ → ((ψ ∨ θ ∨ η) ↔ (χ ∨ τ ∨ ζ)))

Syntax hints:   → wi 2   ↔ wb 127   ∨ wo 195   ∨ w3o 580

This theorem is referenced by:  moeq3 1432  soeq1 2141  solin 2145  dfwe2 2187  weinxp 2467  isowe 2941  f1oweOLD 2944  rdgeq1 2972  rdgeq2 2973  rdglem2 2976  ltsopr 3930  elz 4565

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3or 582

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