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Theorem dp41lemb 1181
 Description: Part of proof (4)=>(1) in Day/Pickering 1982.
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
dp41lemb c2 = ((c2 ∩ ((a0b0) ∪ b1)) ∩ ((a0a1) ∪ b1))

Proof of Theorem dp41lemb
StepHypRef Expression
1 dp41lem.3 . . . . . 6 c2 = ((a0a1) ∩ (b0b1))
2 ancom 74 . . . . . 6 ((a0a1) ∩ (b0b1)) = ((b0b1) ∩ (a0a1))
31, 2tr 62 . . . . 5 c2 = ((b0b1) ∩ (a0a1))
4 leor 159 . . . . . . 7 b0 ≤ (a0b0)
54leror 152 . . . . . 6 (b0b1) ≤ ((a0b0) ∪ b1)
6 leo 158 . . . . . 6 (a0a1) ≤ ((a0a1) ∪ b1)
75, 6le2an 169 . . . . 5 ((b0b1) ∩ (a0a1)) ≤ (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1))
83, 7bltr 138 . . . 4 c2 ≤ (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1))
98df2le2 136 . . 3 (c2 ∩ (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1))) = c2
109cm 61 . 2 c2 = (c2 ∩ (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1)))
11 anass 76 . . 3 ((c2 ∩ ((a0b0) ∪ b1)) ∩ ((a0a1) ∪ b1)) = (c2 ∩ (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1)))
1211cm 61 . 2 (c2 ∩ (((a0b0) ∪ b1) ∩ ((a0a1) ∪ b1))) = ((c2 ∩ ((a0b0) ∪ b1)) ∩ ((a0a1) ∪ b1))
1310, 12tr 62 1 c2 = ((c2 ∩ ((a0b0) ∪ b1)) ∩ ((a0a1) ∪ b1))
 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-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38 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  1192
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