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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 → (ψχ))
bi3d.2 (φ → (θτ))
bi3d.3 (φ → (ηζ))
Assertion
Ref Expression
3orbi123d (φ → ((ψ θ η) ↔ (χ τ ζ)))

Proof of Theorem 3orbi123d
StepHypRef Expression
1 bi3d.1 . . . 4 (φ → (ψχ))
2 bi3d.2 . . . 4 (φ → (θτ))
31, 2orbi12d 690 . . 3 (φ → ((ψ θ) ↔ (χ τ)))
4 bi3d.3 . . 3 (φ → (ηζ))
53, 4orbi12d 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))
85, 6, 73bitr4g 279 1 (φ → ((ψ θ η) ↔ (χ τ ζ)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176   ∨ wo 357   ∨ w3o 933 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8 This theorem depends on definitions:  df-bi 177  df-or 359  df-3or 935 This theorem is referenced by:  moeq3  3013  ltfintri  4466  nncdiv3  6277
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