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Theorem oacom 1011
Description: Commutation law requiring OA. (Contributed by NM, 19-Nov-1998.)
Hypotheses
Ref Expression
oacom.1 d C ((bc) →0 ((a2 b) ∩ (a2 c)))
oacom.2 (d ∩ ((bc) →0 ((a2 b) ∩ (a2 c)))) C (a2 b)
Assertion
Ref Expression
oacom d C ((a2 b) ∩ (a2 c))

Proof of Theorem oacom
StepHypRef Expression
1 oacom.1 . . . . 5 d C ((bc) →0 ((a2 b) ∩ (a2 c)))
21comcom 453 . . . 4 ((bc) →0 ((a2 b) ∩ (a2 c))) C d
3 ancom 74 . . . . . 6 (((bc) →0 ((a2 b) ∩ (a2 c))) ∩ d) = (d ∩ ((bc) →0 ((a2 b) ∩ (a2 c))))
4 oacom.2 . . . . . 6 (d ∩ ((bc) →0 ((a2 b) ∩ (a2 c)))) C (a2 b)
53, 4bctr 181 . . . . 5 (((bc) →0 ((a2 b) ∩ (a2 c))) ∩ d) C (a2 b)
65comcom 453 . . . 4 (a2 b) C (((bc) →0 ((a2 b) ∩ (a2 c))) ∩ d)
72, 6gsth2 490 . . 3 ((a2 b) ∩ ((bc) →0 ((a2 b) ∩ (a2 c)))) C d
87comcom 453 . 2 d C ((a2 b) ∩ ((bc) →0 ((a2 b) ∩ (a2 c))))
9 df-i0 43 . . . 4 ((bc) →0 ((a2 b) ∩ (a2 c))) = ((bc) ∪ ((a2 b) ∩ (a2 c)))
109lan 77 . . 3 ((a2 b) ∩ ((bc) →0 ((a2 b) ∩ (a2 c)))) = ((a2 b) ∩ ((bc) ∪ ((a2 b) ∩ (a2 c))))
11 oath1 1004 . . 3 ((a2 b) ∩ ((bc) ∪ ((a2 b) ∩ (a2 c)))) = ((a2 b) ∩ (a2 c))
1210, 11ax-r2 36 . 2 ((a2 b) ∩ ((bc) →0 ((a2 b) ∩ (a2 c)))) = ((a2 b) ∩ (a2 c))
138, 12cbtr 182 1 d C ((a2 b) ∩ (a2 c))
Colors of variables: term
Syntax hints:   C wc 3   wn 4  wo 6  wa 7  0 wi0 11  2 wi2 13
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  ax-3oa 998
This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i0 43  df-i1 44  df-i2 45  df-le1 130  df-le2 131  df-c1 132  df-c2 133
This theorem is referenced by:  oacom2  1012
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