QLE Home Quantum Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  QLE Home  >  Th. List  >  lem3.3.7i4e1 GIF version

Theorem lem3.3.7i4e1 1069
Description: Equation 3.7 of [PavMeg1999] p. 9. The variable i in the paper is set to 4, and this is the first part of the equation. (Contributed by Roy F. Longton, 28-Jun-2005.) (Revised by Roy F. Longton, 3-Jul-2005.)
Assertion
Ref Expression
lem3.3.7i4e1 (a4 (ab)) = (a4 (ab))

Proof of Theorem lem3.3.7i4e1
StepHypRef Expression
1 lear 161 . . . . . 6 (a ∩ (ab)) ≤ (ab)
2 lea 160 . . . . . . 7 (ab) ≤ a
3 leid 148 . . . . . . 7 (ab) ≤ (ab)
42, 3ler2an 173 . . . . . 6 (ab) ≤ (a ∩ (ab))
51, 4lebi 145 . . . . 5 (a ∩ (ab)) = (ab)
65ax-r5 38 . . . 4 ((a ∩ (ab)) ∪ (a ∩ (ab))) = ((ab) ∪ (a ∩ (ab)))
76ax-r5 38 . . 3 (((a ∩ (ab)) ∪ (a ∩ (ab))) ∪ ((a ∪ (ab)) ∩ (ab) )) = (((ab) ∪ (a ∩ (ab))) ∪ ((a ∪ (ab)) ∩ (ab) ))
82lecon 154 . . . . . . . 8 a ≤ (ab)
98ortha 438 . . . . . . 7 (a ∩ (ab)) = 0
109lor 70 . . . . . 6 ((ab) ∪ (a ∩ (ab))) = ((ab) ∪ 0)
1110ax-r5 38 . . . . 5 (((ab) ∪ (a ∩ (ab))) ∪ ((a ∪ (ab)) ∩ (ab) )) = (((ab) ∪ 0) ∪ ((a ∪ (ab)) ∩ (ab) ))
12 or0 102 . . . . . 6 ((ab) ∪ 0) = (ab)
1312ax-r5 38 . . . . 5 (((ab) ∪ 0) ∪ ((a ∪ (ab)) ∩ (ab) )) = ((ab) ∪ ((a ∪ (ab)) ∩ (ab) ))
14 leor 159 . . . . . . 7 (ab) ≤ (a ∪ (ab))
15 lea 160 . . . . . . 7 ((a ∪ (ab)) ∩ (ab) ) ≤ (a ∪ (ab))
1614, 15lel2or 170 . . . . . 6 ((ab) ∪ ((a ∪ (ab)) ∩ (ab) )) ≤ (a ∪ (ab))
17 leo 158 . . . . . . . . 9 a ≤ (a ∪ (ab))
1817, 8ler2an 173 . . . . . . . 8 a ≤ ((a ∪ (ab)) ∩ (ab) )
1918lerr 150 . . . . . . 7 a ≤ ((ab) ∪ ((a ∪ (ab)) ∩ (ab) ))
20 leo 158 . . . . . . 7 (ab) ≤ ((ab) ∪ ((a ∪ (ab)) ∩ (ab) ))
2119, 20lel2or 170 . . . . . 6 (a ∪ (ab)) ≤ ((ab) ∪ ((a ∪ (ab)) ∩ (ab) ))
2216, 21lebi 145 . . . . 5 ((ab) ∪ ((a ∪ (ab)) ∩ (ab) )) = (a ∪ (ab))
2311, 13, 223tr 65 . . . 4 (((ab) ∪ (a ∩ (ab))) ∪ ((a ∪ (ab)) ∩ (ab) )) = (a ∪ (ab))
24 an1 106 . . . . 5 ((a ∪ (ab)) ∩ 1) = (a ∪ (ab))
2524ax-r1 35 . . . 4 (a ∪ (ab)) = ((a ∪ (ab)) ∩ 1)
263sklem 230 . . . . . 6 ((ab) ∪ (ab)) = 1
2726ax-r1 35 . . . . 5 1 = ((ab) ∪ (ab))
2827lan 77 . . . 4 ((a ∪ (ab)) ∩ 1) = ((a ∪ (ab)) ∩ ((ab) ∪ (ab)))
2923, 25, 283tr 65 . . 3 (((ab) ∪ (a ∩ (ab))) ∪ ((a ∪ (ab)) ∩ (ab) )) = ((a ∪ (ab)) ∩ ((ab) ∪ (ab)))
304, 1lebi 145 . . . . 5 (ab) = (a ∩ (ab))
3130lor 70 . . . 4 ((ab) ∪ (ab)) = ((ab) ∪ (a ∩ (ab)))
3231lan 77 . . 3 ((a ∪ (ab)) ∩ ((ab) ∪ (ab))) = ((a ∪ (ab)) ∩ ((ab) ∪ (a ∩ (ab))))
337, 29, 323tr 65 . 2 (((a ∩ (ab)) ∪ (a ∩ (ab))) ∪ ((a ∪ (ab)) ∩ (ab) )) = ((a ∪ (ab)) ∩ ((ab) ∪ (a ∩ (ab))))
34 df-i4 47 . 2 (a4 (ab)) = (((a ∩ (ab)) ∪ (a ∩ (ab))) ∪ ((a ∪ (ab)) ∩ (ab) ))
35 df-id4 53 . 2 (a4 (ab)) = ((a ∪ (ab)) ∩ ((ab) ∪ (a ∩ (ab))))
3633, 34, 353tr1 63 1 (a4 (ab)) = (a4 (ab))
Colors of variables: term
Syntax hints:   = wb 1   wn 4  wo 6  wa 7  1wt 8  0wf 9  4 wi4 15  4 wid4 21
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
This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i4 47  df-id4 53  df-le1 130  df-le2 131
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator