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Theorem nom30 319
 Description: Part of Lemma 3.3(14) from "Non-Orthomodular Models..." paper. (Contributed by NM, 7-Feb-1999.)
Assertion
Ref Expression
nom30 ((ab) ≡0 a) = (a1 b)

Proof of Theorem nom30
StepHypRef Expression
1 ancom 74 . . 3 (((ab)a) ∩ (a ∪ (ab))) = ((a ∪ (ab)) ∩ ((ab)a))
2 df-id0 49 . . 3 ((ab) ≡0 a) = (((ab)a) ∩ (a ∪ (ab)))
3 df-id0 49 . . 3 (a0 (ab)) = ((a ∪ (ab)) ∩ ((ab)a))
41, 2, 33tr1 63 . 2 ((ab) ≡0 a) = (a0 (ab))
5 nom20 313 . 2 (a0 (ab)) = (a1 b)
64, 5ax-r2 36 1 ((ab) ≡0 a) = (a1 b)
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →1 wi1 12   ≡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-id0 49  df-le1 130  df-le2 131 This theorem is referenced by: (None)
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