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Theorem u1lemnaa 640
 Description: Lemma for Sasaki implication study.
Assertion
Ref Expression
u1lemnaa ((a1 b)a) = (a ∩ (ab ))

Proof of Theorem u1lemnaa
StepHypRef Expression
1 anor2 89 . 2 ((a1 b)a) = ((a1 b) ∪ a )
2 u1lemona 625 . . . 4 ((a1 b) ∪ a ) = (a ∪ (ab))
32ax-r4 37 . . 3 ((a1 b) ∪ a ) = (a ∪ (ab))
4 df-a 40 . . . . 5 (a ∩ (ab )) = (a ∪ (ab ) )
5 df-a 40 . . . . . . . 8 (ab) = (ab )
65lor 70 . . . . . . 7 (a ∪ (ab)) = (a ∪ (ab ) )
76ax-r4 37 . . . . . 6 (a ∪ (ab)) = (a ∪ (ab ) )
87ax-r1 35 . . . . 5 (a ∪ (ab ) ) = (a ∪ (ab))
94, 8ax-r2 36 . . . 4 (a ∩ (ab )) = (a ∪ (ab))
109ax-r1 35 . . 3 (a ∪ (ab)) = (a ∩ (ab ))
113, 10ax-r2 36 . 2 ((a1 b) ∪ a ) = (a ∩ (ab ))
121, 11ax-r2 36 1 ((a1 b)a) = (a ∩ (ab ))
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →1 wi1 12 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 This theorem is referenced by: (None)
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