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Theorem dp15lemg 1158
 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))
dp15lemg.4 c0 = ((a1a2) ∩ (b1b2))
dp15lemg.5 c1 = ((a0a2) ∩ (b0b2))
Assertion
Ref Expression
dp15lemg (((a1a2) ∩ (b1b2)) ∪ (((a0a2) ∩ (b0b2)) ∪ (b1 ∩ (a0a1)))) = ((c0c1) ∪ (b1 ∩ (a0a1)))

Proof of Theorem dp15lemg
StepHypRef Expression
1 dp15lemg.4 . . . 4 c0 = ((a1a2) ∩ (b1b2))
2 dp15lemg.5 . . . . 5 c1 = ((a0a2) ∩ (b0b2))
32ror 71 . . . 4 (c1 ∪ (b1 ∩ (a0a1))) = (((a0a2) ∩ (b0b2)) ∪ (b1 ∩ (a0a1)))
41, 32or 72 . . 3 (c0 ∪ (c1 ∪ (b1 ∩ (a0a1)))) = (((a1a2) ∩ (b1b2)) ∪ (((a0a2) ∩ (b0b2)) ∪ (b1 ∩ (a0a1))))
54cm 61 . 2 (((a1a2) ∩ (b1b2)) ∪ (((a0a2) ∩ (b0b2)) ∪ (b1 ∩ (a0a1)))) = (c0 ∪ (c1 ∪ (b1 ∩ (a0a1))))
6 orass 75 . . 3 ((c0c1) ∪ (b1 ∩ (a0a1))) = (c0 ∪ (c1 ∪ (b1 ∩ (a0a1))))
76cm 61 . 2 (c0 ∪ (c1 ∪ (b1 ∩ (a0a1)))) = ((c0c1) ∪ (b1 ∩ (a0a1)))
85, 7tr 62 1 (((a1a2) ∩ (b1b2)) ∪ (((a0a2) ∩ (b0b2)) ∪ (b1 ∩ (a0a1)))) = ((c0c1) ∪ (b1 ∩ (a0a1)))
 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-r1 35  ax-r2 36  ax-r5 38 This theorem is referenced by:  dp15lemh  1159
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