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Theorem dp15lemc 1154
 Description: Part of proof (1)=>(5) in Day/Pickering 1982.
Hypotheses
Ref Expression
dp15lema.1 d = (a2 ∪ (a0 ∩ (a1b1)))
dp15lema.2 p0 = ((a1b1) ∩ (a2b2))
dp15lema.3 e = (b0 ∩ (a0p0))
Assertion
Ref Expression
dp15lemc ((a0a1) ∩ ((b0 ∩ (a0p0)) ∪ b1)) ≤ (((a0 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ ((b0 ∩ (a0p0)) ∪ b2)) ∪ ((a1 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ (b1b2)))

Proof of Theorem dp15lemc
StepHypRef Expression
1 dp15lema.1 . . 3 d = (a2 ∪ (a0 ∩ (a1b1)))
2 dp15lema.2 . . 3 p0 = ((a1b1) ∩ (a2b2))
3 dp15lema.3 . . 3 e = (b0 ∩ (a0p0))
41, 2, 3dp15lemb 1153 . 2 ((a0a1) ∩ (eb1)) ≤ (((a0d) ∩ (eb2)) ∪ ((a1d) ∩ (b1b2)))
53ror 71 . . 3 (eb1) = ((b0 ∩ (a0p0)) ∪ b1)
65lan 77 . 2 ((a0a1) ∩ (eb1)) = ((a0a1) ∩ ((b0 ∩ (a0p0)) ∪ b1))
71lor 70 . . . 4 (a0d) = (a0 ∪ (a2 ∪ (a0 ∩ (a1b1))))
83ror 71 . . . 4 (eb2) = ((b0 ∩ (a0p0)) ∪ b2)
97, 82an 79 . . 3 ((a0d) ∩ (eb2)) = ((a0 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ ((b0 ∩ (a0p0)) ∪ b2))
101lor 70 . . . 4 (a1d) = (a1 ∪ (a2 ∪ (a0 ∩ (a1b1))))
1110ran 78 . . 3 ((a1d) ∩ (b1b2)) = ((a1 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ (b1b2))
129, 112or 72 . 2 (((a0d) ∩ (eb2)) ∪ ((a1d) ∩ (b1b2))) = (((a0 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ ((b0 ∩ (a0p0)) ∪ b2)) ∪ ((a1 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ (b1b2)))
134, 6, 12le3tr2 141 1 ((a0a1) ∩ ((b0 ∩ (a0p0)) ∪ b1)) ≤ (((a0 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ ((b0 ∩ (a0p0)) ∪ b2)) ∪ ((a1 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ (b1b2)))
 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:  dp15lemh  1159
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