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Theorem nom50 331
 Description: Part of Lemma 3.3(15) from "Non-Orthomodular Models..." paper. (Contributed by NM, 7-Feb-1999.)
Assertion
Ref Expression
nom50 ((ab) ≡0 b) = (a2 b)

Proof of Theorem nom50
StepHypRef Expression
1 ancom 74 . . . . . . . 8 (ba ) = (ab )
2 anor3 90 . . . . . . . 8 (ab ) = (ab)
31, 2ax-r2 36 . . . . . . 7 (ba ) = (ab)
43lor 70 . . . . . 6 (b ∪ (ba )) = (b ∪ (ab) )
53ax-r4 37 . . . . . . 7 (ba ) = (ab)
65ax-r5 38 . . . . . 6 ((ba )b ) = ((ab) b )
74, 62an 79 . . . . 5 ((b ∪ (ba )) ∩ ((ba )b )) = ((b ∪ (ab) ) ∩ ((ab) b ))
87ax-r1 35 . . . 4 ((b ∪ (ab) ) ∩ ((ab) b )) = ((b ∪ (ba )) ∩ ((ba )b ))
9 df-id0 49 . . . 4 (b0 (ab) ) = ((b ∪ (ab) ) ∩ ((ab) b ))
10 df-id0 49 . . . 4 (b0 (ba )) = ((b ∪ (ba )) ∩ ((ba )b ))
118, 9, 103tr1 63 . . 3 (b0 (ab) ) = (b0 (ba ))
12 nom20 313 . . 3 (b0 (ba )) = (b1 a )
1311, 12ax-r2 36 . 2 (b0 (ab) ) = (b1 a )
14 nomcon0 301 . 2 ((ab) ≡0 b) = (b0 (ab) )
15 i2i1 267 . 2 (a2 b) = (b1 a )
1613, 14, 153tr1 63 1 ((ab) ≡0 b) = (a2 b)
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →1 wi1 12   →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:  nom60  337
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