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Theorem nomcon0 293
Description: Lemma for "Non-Orthomodular Models..." paper.
Assertion
Ref Expression
nomcon0 (a0 b) = (b0 a )

Proof of Theorem nomcon0
StepHypRef Expression
1 ax-a2 30 . . . 4 (ab) = (ba )
2 ax-a1 29 . . . . 5 b = b
32ax-r5 37 . . . 4 (ba ) = (b a )
41, 3ax-r2 35 . . 3 (ab) = (b a )
5 ax-a2 30 . . . 4 (ba) = (ab )
6 ax-a1 29 . . . . 5 a = a
76ax-r5 37 . . . 4 (ab ) = (a b )
85, 7ax-r2 35 . . 3 (ba) = (a b )
94, 82an 72 . 2 ((ab) ∩ (ba)) = ((b a ) ∩ (a b ))
10 df-id0 48 . 2 (a0 b) = ((ab) ∩ (ba))
11 df-id0 48 . 2 (b0 a ) = ((b a ) ∩ (a b ))
129, 10, 113tr1 60 1 (a0 b) = (b0 a )
Colors of variables: term
Syntax hints:   = wb 1   wn 4   ∪ wo 6   ∩ wa 7   ≡0 wid0 18
This theorem is referenced by:  nom50 323
This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37
This theorem depends on definitions:  df-a 39  df-id0 48
metamath.org