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Theorem dp41lemd 1186
Description: Part of proof (4)=>(1) in Day/Pickering 1982. (Contributed by NM, 3-Apr-2012.)
Hypotheses
Ref Expression
dp41lem.1 c0 = ((a1a2) ∩ (b1b2))
dp41lem.2 c1 = ((a0a2) ∩ (b0b2))
dp41lem.3 c2 = ((a0a1) ∩ (b0b1))
dp41lem.4 p = (((a0b0) ∩ (a1b1)) ∩ (a2b2))
dp41lem.5 p2 = ((a0b0) ∩ (a1b1))
dp41lem.6 p2 ≤ (a2b2)
Assertion
Ref Expression
dp41lemd (c2 ∩ ((a0b1) ∪ (c2 ∩ (c0c1)))) = (c2 ∩ ((c0c1) ∪ (c2 ∩ (a0b1))))

Proof of Theorem dp41lemd
StepHypRef Expression
1 mldual 1124 . 2 (c2 ∩ ((a0b1) ∪ (c2 ∩ (c0c1)))) = ((c2 ∩ (a0b1)) ∪ (c2 ∩ (c0c1)))
2 ancom 74 . . 3 (c2 ∩ (c0c1)) = ((c0c1) ∩ c2)
32lor 70 . 2 ((c2 ∩ (a0b1)) ∪ (c2 ∩ (c0c1))) = ((c2 ∩ (a0b1)) ∪ ((c0c1) ∩ c2))
4 lea 160 . . . 4 (c2 ∩ (a0b1)) ≤ c2
54ml2i 1125 . . 3 ((c2 ∩ (a0b1)) ∪ ((c0c1) ∩ c2)) = (((c2 ∩ (a0b1)) ∪ (c0c1)) ∩ c2)
6 ancom 74 . . 3 (((c2 ∩ (a0b1)) ∪ (c0c1)) ∩ c2) = (c2 ∩ ((c2 ∩ (a0b1)) ∪ (c0c1)))
7 ax-a2 31 . . . 4 ((c2 ∩ (a0b1)) ∪ (c0c1)) = ((c0c1) ∪ (c2 ∩ (a0b1)))
87lan 77 . . 3 (c2 ∩ ((c2 ∩ (a0b1)) ∪ (c0c1))) = (c2 ∩ ((c0c1) ∪ (c2 ∩ (a0b1))))
95, 6, 83tr 65 . 2 ((c2 ∩ (a0b1)) ∪ ((c0c1) ∩ c2)) = (c2 ∩ ((c0c1) ∪ (c2 ∩ (a0b1))))
101, 3, 93tr 65 1 (c2 ∩ ((a0b1) ∪ (c2 ∩ (c0c1)))) = (c2 ∩ ((c0c1) ∪ (c2 ∩ (a0b1))))
Colors of variables: term
Syntax hints:   = wb 1  wle 2  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 1122
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:  dp41lemm  1194
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