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Theorem nom60 337
 Description: Part of Lemma 3.3(15) from "Non-Orthomodular Models..." paper.
Assertion
Ref Expression
nom60 (b0 (ab)) = (a2 b)

Proof of Theorem nom60
StepHypRef Expression
1 ancom 74 . . 3 ((b ∪ (ab)) ∩ ((ab)b)) = (((ab)b) ∩ (b ∪ (ab)))
2 df-id0 49 . . 3 (b0 (ab)) = ((b ∪ (ab)) ∩ ((ab)b))
3 df-id0 49 . . 3 ((ab) ≡0 b) = (((ab)b) ∩ (b ∪ (ab)))
41, 2, 33tr1 63 . 2 (b0 (ab)) = ((ab) ≡0 b)
5 nom50 331 . 2 ((ab) ≡0 b) = (a2 b)
64, 5ax-r2 36 1 (b0 (ab)) = (a2 b)
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →2 wi2 13   ≡0 wid0 17 This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38 This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44  df-i2 45  df-id0 49  df-le1 130  df-le2 131 This theorem is referenced by: (None)
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