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Theorem nom55 336
 Description: Part of Lemma 3.3(15) from "Non-Orthomodular Models..." paper.
Assertion
Ref Expression
nom55 ((ab) ≡ b) = (a2 b)

Proof of Theorem nom55
StepHypRef Expression
1 nom25 318 . 2 (b ≡ (ba )) = (b1 a )
2 conb 122 . . 3 ((ab) ≡ b) = ((ab)b )
3 bicom 96 . . 3 ((ab)b ) = (b ≡ (ab) )
4 ancom 74 . . . . . 6 (ba ) = (ab )
5 anor3 90 . . . . . 6 (ab ) = (ab)
64, 5ax-r2 36 . . . . 5 (ba ) = (ab)
76ax-r1 35 . . . 4 (ab) = (ba )
87lbi 97 . . 3 (b ≡ (ab) ) = (b ≡ (ba ))
92, 3, 83tr 65 . 2 ((ab) ≡ b) = (b ≡ (ba ))
10 i2i1 267 . 2 (a2 b) = (b1 a )
111, 9, 103tr1 63 1 ((ab) ≡ b) = (a2 b)
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7   →1 wi1 12   →2 wi2 13 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-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-i2 45 This theorem is referenced by:  nom65  342
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