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Theorem 3orbi123d 1205
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 706 . . 3 (φ → ((ψ θ) ↔ (χ τ)))
4 bi3d.3 . . 3 (φ → (ηζ))
53, 4orbi12d 706 . 2 (φ → (((ψ θ) η) ↔ ((χ τ) ζ)))
6 df-3or 885 . 2 ((ψ θ η) ↔ ((ψ θ) η))
7 df-3or 885 . 2 ((χ τ ζ) ↔ ((χ τ) ζ))
85, 6, 73bitr4g 212 1 (φ → ((ψ θ η) ↔ (χ τ ζ)))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98   wo 628   w3o 883
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 629
This theorem depends on definitions:  df-bi 110  df-3or 885
This theorem is referenced by:  ordtriexmid  4210  nntri3or  6011  ltsopi  6304  pitri3or  6306  nqtri3or  6380  elz  7983  ztri3or  8024
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