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Theorem dp35lembb 1175
 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
dp35lembb (b0 ∩ (a0p0)) ≤ (b0 ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))

Proof of Theorem dp35lembb
StepHypRef Expression
1 dp35lem.1 . . 3 c0 = ((a1a2) ∩ (b1b2))
2 dp35lem.2 . . 3 c1 = ((a0a2) ∩ (b0b2))
3 dp35lem.3 . . 3 c2 = ((a0a1) ∩ (b0b1))
4 dp35lem.4 . . 3 p0 = ((a1b1) ∩ (a2b2))
5 dp35lem.5 . . 3 p = (((a0b0) ∩ (a1b1)) ∩ (a2b2))
61, 2, 3, 4, 5dp35lemd 1172 . 2 (b0 ∩ (a0p0)) ≤ (b0 ∩ (((a0b0) ∪ b1) ∪ (c2 ∩ (c0c1))))
71, 2, 3, 4, 5dp35lemc 1173 . . 3 (b0 ∩ (((a0b0) ∪ b1) ∪ (c2 ∩ (c0c1)))) = (b0 ∩ (b1 ∪ (c2 ∩ (c0c1))))
81, 2, 3, 4, 5dp35lemb 1174 . . 3 (b0 ∩ (b1 ∪ (c2 ∩ (c0c1)))) = (b0 ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))
97, 8tr 62 . 2 (b0 ∩ (((a0b0) ∪ b1) ∪ (c2 ∩ (c0c1)))) = (b0 ∩ (b1 ∪ ((a0a1) ∩ (c0c1))))
106, 9lbtr 139 1 (b0 ∩ (a0p0)) ≤ (b0 ∩ (b1 ∪ ((a0a1) ∩ (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 1120  ax-arg 1151 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:  dp35lema  1176
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