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Theorem ud3lem3b 573
 Description: Lemma for unified disjunction. (Contributed by NM, 27-Nov-1997.)
Assertion
Ref Expression
ud3lem3b ((a3 b) ∩ (ab) ) = 0

Proof of Theorem ud3lem3b
StepHypRef Expression
1 ud3lem0c 279 . . 3 (a3 b) = (((ab ) ∩ (ab)) ∩ (a ∪ (ab )))
21ran 78 . 2 ((a3 b) ∩ (ab) ) = ((((ab ) ∩ (ab)) ∩ (a ∪ (ab ))) ∩ (ab) )
3 an32 83 . . 3 ((((ab ) ∩ (ab)) ∩ (a ∪ (ab ))) ∩ (ab) ) = ((((ab ) ∩ (ab)) ∩ (ab) ) ∩ (a ∪ (ab )))
4 anass 76 . . . . . 6 (((ab ) ∩ (ab)) ∩ (ab) ) = ((ab ) ∩ ((ab) ∩ (ab) ))
5 dff 101 . . . . . . . . 9 0 = ((ab) ∩ (ab) )
65ax-r1 35 . . . . . . . 8 ((ab) ∩ (ab) ) = 0
76lan 77 . . . . . . 7 ((ab ) ∩ ((ab) ∩ (ab) )) = ((ab ) ∩ 0)
8 an0 108 . . . . . . 7 ((ab ) ∩ 0) = 0
97, 8ax-r2 36 . . . . . 6 ((ab ) ∩ ((ab) ∩ (ab) )) = 0
104, 9ax-r2 36 . . . . 5 (((ab ) ∩ (ab)) ∩ (ab) ) = 0
1110ran 78 . . . 4 ((((ab ) ∩ (ab)) ∩ (ab) ) ∩ (a ∪ (ab ))) = (0 ∩ (a ∪ (ab )))
12 an0r 109 . . . 4 (0 ∩ (a ∪ (ab ))) = 0
1311, 12ax-r2 36 . . 3 ((((ab ) ∩ (ab)) ∩ (ab) ) ∩ (a ∪ (ab ))) = 0
143, 13ax-r2 36 . 2 ((((ab ) ∩ (ab)) ∩ (a ∪ (ab ))) ∩ (ab) ) = 0
152, 14ax-r2 36 1 ((a3 b) ∩ (ab) ) = 0
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →3 wi3 14 This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  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-i3 46 This theorem is referenced by:  ud3lem3  576
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