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Theorem dp15lemd 1157
 Description: Part of proof (1)=>(5) in Day/Pickering 1982. (Contributed by NM, 1-Apr-2012.)
Hypotheses
Ref Expression
dp15lema.1 d = (a2 ∪ (a0 ∩ (a1b1)))
dp15lema.2 p0 = ((a1b1) ∩ (a2b2))
dp15lema.3 e = (b0 ∩ (a0p0))
Assertion
Ref Expression
dp15lemd (((a0 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ ((b0 ∩ (a0p0)) ∪ b2)) ∪ ((a1 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ (b1b2))) = (((a0a2) ∩ ((b0 ∩ (a0p0)) ∪ b2)) ∪ (((a1a2) ∪ (a0 ∩ (a1b1))) ∩ (b1b2)))

Proof of Theorem dp15lemd
StepHypRef Expression
1 or12 80 . . . 4 (a0 ∪ (a2 ∪ (a0 ∩ (a1b1)))) = (a2 ∪ (a0 ∪ (a0 ∩ (a1b1))))
2 orabs 120 . . . . 5 (a0 ∪ (a0 ∩ (a1b1))) = a0
32lor 70 . . . 4 (a2 ∪ (a0 ∪ (a0 ∩ (a1b1)))) = (a2a0)
4 orcom 73 . . . 4 (a2a0) = (a0a2)
51, 3, 43tr 65 . . 3 (a0 ∪ (a2 ∪ (a0 ∩ (a1b1)))) = (a0a2)
65ran 78 . 2 ((a0 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ ((b0 ∩ (a0p0)) ∪ b2)) = ((a0a2) ∩ ((b0 ∩ (a0p0)) ∪ b2))
7 orass 75 . . . 4 ((a1a2) ∪ (a0 ∩ (a1b1))) = (a1 ∪ (a2 ∪ (a0 ∩ (a1b1))))
87ran 78 . . 3 (((a1a2) ∪ (a0 ∩ (a1b1))) ∩ (b1b2)) = ((a1 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ (b1b2))
98cm 61 . 2 ((a1 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ (b1b2)) = (((a1a2) ∪ (a0 ∩ (a1b1))) ∩ (b1b2))
106, 92or 72 1 (((a0 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ ((b0 ∩ (a0p0)) ∪ b2)) ∪ ((a1 ∪ (a2 ∪ (a0 ∩ (a1b1)))) ∩ (b1b2))) = (((a0a2) ∩ ((b0 ∩ (a0p0)) ∪ b2)) ∪ (((a1a2) ∪ (a0 ∩ (a1b1))) ∩ (b1b2)))
 Colors of variables: term Syntax hints:   = wb 1   ∪ wo 6   ∩ wa 7 This theorem was proved from axioms:  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38 This theorem depends on definitions:  df-a 40 This theorem is referenced by:  dp15lemh  1161
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