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Theorem nomcon0 301
 Description: Lemma for "Non-Orthomodular Models..." paper.
Assertion
Ref Expression
nomcon0 (a0 b) = (b0 a )

Proof of Theorem nomcon0
StepHypRef Expression
1 ax-a2 31 . . . 4 (ab) = (ba )
2 ax-a1 30 . . . . 5 b = b
32ax-r5 38 . . . 4 (ba ) = (b a )
41, 3ax-r2 36 . . 3 (ab) = (b a )
5 ax-a2 31 . . . 4 (ba) = (ab )
6 ax-a1 30 . . . . 5 a = a
76ax-r5 38 . . . 4 (ab ) = (a b )
85, 7ax-r2 36 . . 3 (ba) = (a b )
94, 82an 79 . 2 ((ab) ∩ (ba)) = ((b a ) ∩ (a b ))
10 df-id0 49 . 2 (a0 b) = ((ab) ∩ (ba))
11 df-id0 49 . 2 (b0 a ) = ((b a ) ∩ (a b ))
129, 10, 113tr1 63 1 (a0 b) = (b0 a )
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   ≡0 wid0 17 This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38 This theorem depends on definitions:  df-a 40  df-id0 49 This theorem is referenced by:  nom50  331
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