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Theorem u1lemab 610
 Description: Lemma for Sasaki implication study. Equation 4.10 of [MegPav2000] p. 23. This is the second part of the equation. (Contributed by NM, 14-Dec-1997.)
Assertion
Ref Expression
u1lemab ((a1 b) ∩ b) = ((ab) ∪ (ab))

Proof of Theorem u1lemab
StepHypRef Expression
1 df-i1 44 . . 3 (a1 b) = (a ∪ (ab))
21ran 78 . 2 ((a1 b) ∩ b) = ((a ∪ (ab)) ∩ b)
3 ax-a2 31 . . . . 5 (a ∪ (ab)) = ((ab) ∪ a )
43ran 78 . . . 4 ((a ∪ (ab)) ∩ b) = (((ab) ∪ a ) ∩ b)
5 coman2 186 . . . . 5 (ab) C b
6 coman1 185 . . . . . 6 (ab) C a
76comcom2 183 . . . . 5 (ab) C a
85, 7fh2r 474 . . . 4 (((ab) ∪ a ) ∩ b) = (((ab) ∩ b) ∪ (ab))
94, 8ax-r2 36 . . 3 ((a ∪ (ab)) ∩ b) = (((ab) ∩ b) ∪ (ab))
10 anass 76 . . . . 5 ((ab) ∩ b) = (a ∩ (bb))
11 anidm 111 . . . . . 6 (bb) = b
1211lan 77 . . . . 5 (a ∩ (bb)) = (ab)
1310, 12ax-r2 36 . . . 4 ((ab) ∩ b) = (ab)
1413ax-r5 38 . . 3 (((ab) ∩ b) ∪ (ab)) = ((ab) ∪ (ab))
159, 14ax-r2 36 . 2 ((a ∪ (ab)) ∩ b) = ((ab) ∪ (ab))
162, 15ax-r2 36 1 ((a1 b) ∩ b) = ((ab) ∪ (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-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439 This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133 This theorem is referenced by:  u1lemnonb  675  i1com  708  u1lem3  749  u1lem11  780  sadm3  838  negantlem2  849  negantlem3  850  negantlem10  861  neg3antlem1  864  oa4to4u  973  oa3-6lem  980  oa3-u1lem  985  oa3-u2lem  986  oa3-1to5  993  lem4.6.2e2  1083
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