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Theorem u1lemaa 600
 Description: Lemma for Sasaki implication study. Equation 4.10 of [MegPav2000] p. 23. This is the second part of the equation.
Assertion
Ref Expression
u1lemaa ((a1 b) ∩ a) = (ab)

Proof of Theorem u1lemaa
StepHypRef Expression
1 df-i1 44 . . 3 (a1 b) = (a ∪ (ab))
21ran 78 . 2 ((a1 b) ∩ a) = ((a ∪ (ab)) ∩ a)
3 comid 187 . . . . 5 a C a
43comcom2 183 . . . 4 a C a
5 comanr1 464 . . . 4 a C (ab)
64, 5fh1r 473 . . 3 ((a ∪ (ab)) ∩ a) = ((aa) ∪ ((ab) ∩ a))
7 ax-a2 31 . . . . 5 ((aa) ∪ ((ab) ∩ a)) = (((ab) ∩ a) ∪ (aa))
8 an32 83 . . . . . . 7 ((ab) ∩ a) = ((aa) ∩ b)
9 anidm 111 . . . . . . . 8 (aa) = a
109ran 78 . . . . . . 7 ((aa) ∩ b) = (ab)
118, 10ax-r2 36 . . . . . 6 ((ab) ∩ a) = (ab)
12 ancom 74 . . . . . . 7 (aa) = (aa )
13 dff 101 . . . . . . . 8 0 = (aa )
1413ax-r1 35 . . . . . . 7 (aa ) = 0
1512, 14ax-r2 36 . . . . . 6 (aa) = 0
1611, 152or 72 . . . . 5 (((ab) ∩ a) ∪ (aa)) = ((ab) ∪ 0)
177, 16ax-r2 36 . . . 4 ((aa) ∪ ((ab) ∩ a)) = ((ab) ∪ 0)
18 or0 102 . . . 4 ((ab) ∪ 0) = (ab)
1917, 18ax-r2 36 . . 3 ((aa) ∪ ((ab) ∩ a)) = (ab)
206, 19ax-r2 36 . 2 ((a ∪ (ab)) ∩ a) = (ab)
212, 20ax-r2 36 1 ((a1 b) ∩ a) = (ab)
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →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:  u1lemnona  665  u12lembi  726  u1lem5  761  negantlem2  849  kb10iii  893  oas  925  oau  929  oaur  930  oa6to4  958  oa8to5  972
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