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Theorem dp41lemc 1185
 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
dp41lemc ((c2 ∩ ((a0b0) ∪ b1)) ∩ ((a0a1) ∪ b1)) ≤ (c2 ∩ ((a0b1) ∪ (c2 ∩ (c0c1))))

Proof of Theorem dp41lemc
StepHypRef Expression
1 anass 76 . 2 ((c2 ∩ ((a0b0) ∪ b1)) ∩ ((a0a1) ∪ b1)) = (c2 ∩ (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1)))
2 dp41lem.1 . . . . 5 c0 = ((a1a2) ∩ (b1b2))
3 dp41lem.2 . . . . 5 c1 = ((a0a2) ∩ (b0b2))
4 dp41lem.3 . . . . 5 c2 = ((a0a1) ∩ (b0b1))
5 dp41lem.4 . . . . 5 p = (((a0b0) ∩ (a1b1)) ∩ (a2b2))
6 dp41lem.5 . . . . 5 p2 = ((a0b0) ∩ (a1b1))
7 dp41lem.6 . . . . 5 p2 ≤ (a2b2)
82, 3, 4, 5, 6, 7dp41lemc0 1184 . . . 4 (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1)) = ((a0b1) ∪ ((a0b0) ∩ (a1b1)))
9 leo 158 . . . . 5 (a0b1) ≤ ((a0b1) ∪ (c2 ∩ (c0c1)))
102, 3, 4, 5, 6, 7dp41lema 1182 . . . . 5 ((a0b0) ∩ (a1b1)) ≤ ((a0b1) ∪ (c2 ∩ (c0c1)))
119, 10lel2or 170 . . . 4 ((a0b1) ∪ ((a0b0) ∩ (a1b1))) ≤ ((a0b1) ∪ (c2 ∩ (c0c1)))
128, 11bltr 138 . . 3 (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1)) ≤ ((a0b1) ∪ (c2 ∩ (c0c1)))
1312lelan 167 . 2 (c2 ∩ (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1))) ≤ (c2 ∩ ((a0b1) ∪ (c2 ∩ (c0c1))))
141, 13bltr 138 1 ((c2 ∩ ((a0b0) ∪ b1)) ∩ ((a0a1) ∪ b1)) ≤ (c2 ∩ ((a0b1) ∪ (c2 ∩ (c0c1))))
 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-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-ml 1122  ax-arg 1153 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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