Quantum Logic Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  QLE Home  >  Th. List  >  lem4.6.6i1j0 GIF version

Theorem lem4.6.6i1j0 1090
 Description: Equation 4.14 of [MegPav2000] p. 23. The variable i in the paper is set to 1, and j is set to 0. (Contributed by Roy F. Longton, 3-Jul-05.)
Assertion
Ref Expression
lem4.6.6i1j0 ((a1 b) ∪ (a0 b)) = (a0 b)

Proof of Theorem lem4.6.6i1j0
StepHypRef Expression
1 lear 161 . . . 4 (ab) ≤ b
21lelor 166 . . 3 (a ∪ (ab)) ≤ (ab)
32df-le2 131 . 2 ((a ∪ (ab)) ∪ (ab)) = (ab)
4 df-i1 44 . . 3 (a1 b) = (a ∪ (ab))
5 df-i0 43 . . 3 (a0 b) = (ab)
64, 52or 72 . 2 ((a1 b) ∪ (a0 b)) = ((a ∪ (ab)) ∪ (ab))
73, 6, 53tr1 63 1 ((a1 b) ∪ (a0 b)) = (a0 b)
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →0 wi0 11   →1 wi1 12 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-i0 43  df-i1 44  df-le1 130  df-le2 131 This theorem is referenced by: (None)
 Copyright terms: Public domain W3C validator