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Theorem 3or6 1201
Description: Analog of or4 675 for triple conjunction. (Contributed by Scott Fenton, 16-Mar-2011.)
Assertion
Ref Expression
3or6 (((φ ψ) (χ θ) (τ η)) ↔ ((φ χ τ) (ψ θ η)))

Proof of Theorem 3or6
StepHypRef Expression
1 or4 675 . . 3 ((((φ χ) τ) ((ψ θ) η)) ↔ (((φ χ) (ψ θ)) (τ η)))
2 or4 675 . . . 4 (((φ χ) (ψ θ)) ↔ ((φ ψ) (χ θ)))
32orbi1i 667 . . 3 ((((φ χ) (ψ θ)) (τ η)) ↔ (((φ ψ) (χ θ)) (τ η)))
41, 3bitr2i 174 . 2 ((((φ ψ) (χ θ)) (τ η)) ↔ (((φ χ) τ) ((ψ θ) η)))
5 df-3or 872 . 2 (((φ ψ) (χ θ) (τ η)) ↔ (((φ ψ) (χ θ)) (τ η)))
6 df-3or 872 . . 3 ((φ χ τ) ↔ ((φ χ) τ))
7 df-3or 872 . . 3 ((ψ θ η) ↔ ((ψ θ) η))
86, 7orbi12i 668 . 2 (((φ χ τ) (ψ θ η)) ↔ (((φ χ) τ) ((ψ θ) η)))
94, 5, 83bitr4i 201 1 (((φ ψ) (χ θ) (τ η)) ↔ ((φ χ τ) (ψ θ η)))
Colors of variables: wff set class
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)
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