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Theorem 3or6 1263
 Description: Analog of or4 514 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 514 . . 3 ((((φ χ) τ) ((ψ θ) η)) ↔ (((φ χ) (ψ θ)) (τ η)))
2 or4 514 . . . 4 (((φ χ) (ψ θ)) ↔ ((φ ψ) (χ θ)))
32orbi1i 506 . . 3 ((((φ χ) (ψ θ)) (τ η)) ↔ (((φ ψ) (χ θ)) (τ η)))
41, 3bitr2i 241 . 2 ((((φ ψ) (χ θ)) (τ η)) ↔ (((φ χ) τ) ((ψ θ) η)))
5 df-3or 935 . 2 (((φ ψ) (χ θ) (τ η)) ↔ (((φ ψ) (χ θ)) (τ η)))
6 df-3or 935 . . 3 ((φ χ τ) ↔ ((φ χ) τ))
7 df-3or 935 . . 3 ((ψ θ η) ↔ ((ψ θ) η))
86, 7orbi12i 507 . 2 (((φ χ τ) (ψ θ η)) ↔ (((φ χ) τ) ((ψ θ) η)))
94, 5, 83bitr4i 268 1 (((φ ψ) (χ θ) (τ η)) ↔ ((φ χ τ) (ψ θ η)))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 176   ∨ wo 357   ∨ w3o 933 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8 This theorem depends on definitions:  df-bi 177  df-or 359  df-3or 935 This theorem is referenced by: (None)
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