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Theorem dp35lemb 1174
 Description: Part of proof (3)=>(5) in Day/Pickering 1982.
Hypotheses
Ref Expression
dp35lem.1 c0 = ((a1a2) ∩ (b1b2))
dp35lem.2 c1 = ((a0a2) ∩ (b0b2))
dp35lem.3 c2 = ((a0a1) ∩ (b0b1))
dp35lem.4 p0 = ((a1b1) ∩ (a2b2))
dp35lem.5 p = (((a0b0) ∩ (a1b1)) ∩ (a2b2))
Assertion
Ref Expression
dp35lemb (b0 ∩ (b1 ∪ (c2 ∩ (c0c1)))) = (b0 ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))

Proof of Theorem dp35lemb
StepHypRef Expression
1 dp35lem.3 . . . . . . 7 c2 = ((a0a1) ∩ (b0b1))
21ran 78 . . . . . 6 (c2 ∩ (c0c1)) = (((a0a1) ∩ (b0b1)) ∩ (c0c1))
3 an32 83 . . . . . 6 (((a0a1) ∩ (b0b1)) ∩ (c0c1)) = (((a0a1) ∩ (c0c1)) ∩ (b0b1))
42, 3tr 62 . . . . 5 (c2 ∩ (c0c1)) = (((a0a1) ∩ (c0c1)) ∩ (b0b1))
54lor 70 . . . 4 (b1 ∪ (c2 ∩ (c0c1))) = (b1 ∪ (((a0a1) ∩ (c0c1)) ∩ (b0b1)))
6 leor 159 . . . . 5 b1 ≤ (b0b1)
76ml2i 1123 . . . 4 (b1 ∪ (((a0a1) ∩ (c0c1)) ∩ (b0b1))) = ((b1 ∪ ((a0a1) ∩ (c0c1))) ∩ (b0b1))
8 ancom 74 . . . 4 ((b1 ∪ ((a0a1) ∩ (c0c1))) ∩ (b0b1)) = ((b0b1) ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))
95, 7, 83tr 65 . . 3 (b1 ∪ (c2 ∩ (c0c1))) = ((b0b1) ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))
109lan 77 . 2 (b0 ∩ (b1 ∪ (c2 ∩ (c0c1)))) = (b0 ∩ ((b0b1) ∩ (b1 ∪ ((a0a1) ∩ (c0c1)))))
11 anass 76 . . 3 ((b0 ∩ (b0b1)) ∩ (b1 ∪ ((a0a1) ∩ (c0c1)))) = (b0 ∩ ((b0b1) ∩ (b1 ∪ ((a0a1) ∩ (c0c1)))))
1211cm 61 . 2 (b0 ∩ ((b0b1) ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))) = ((b0 ∩ (b0b1)) ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))
13 anabs 121 . . 3 (b0 ∩ (b0b1)) = b0
1413ran 78 . 2 ((b0 ∩ (b0b1)) ∩ (b1 ∪ ((a0a1) ∩ (c0c1)))) = (b0 ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))
1510, 12, 143tr 65 1 (b0 ∩ (b1 ∪ (c2 ∩ (c0c1)))) = (b0 ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))
 Colors of variables: term Syntax hints:   = wb 1   ∪ wo 6   ∩ wa 7 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  ax-ml 1120 This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131 This theorem is referenced by:  dp35lembb  1175
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