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Theorem lem4.6.6i0j1 1086
 Description: Equation 4.14 of [MegPav2000] p. 23. The variable i in the paper is set to 0, and j is set to 1. (Contributed by Roy F. Longton, 3-Jul-05.)
Assertion
Ref Expression
lem4.6.6i0j1 ((a0 b) ∪ (a1 b)) = (a0 b)

Proof of Theorem lem4.6.6i0j1
StepHypRef Expression
1 leid 148 . . . 4 (ab) ≤ (ab)
2 lear 161 . . . . 5 (ab) ≤ b
32lelor 166 . . . 4 (a ∪ (ab)) ≤ (ab)
41, 3lel2or 170 . . 3 ((ab) ∪ (a ∪ (ab))) ≤ (ab)
5 leo 158 . . 3 (ab) ≤ ((ab) ∪ (a ∪ (ab)))
64, 5lebi 145 . 2 ((ab) ∪ (a ∪ (ab))) = (ab)
7 df-i0 43 . . 3 (a0 b) = (ab)
8 df-i1 44 . . 3 (a1 b) = (a ∪ (ab))
97, 82or 72 . 2 ((a0 b) ∪ (a1 b)) = ((ab) ∪ (a ∪ (ab)))
106, 9, 73tr1 63 1 ((a0 b) ∪ (a1 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)
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