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Theorem lem3.3.3 1052
 Description: Equation 3.3 of [PavMeg1999] p. 9. (Contributed by Roy F. Longton, 3-Jul-05.)
Assertion
Ref Expression
lem3.3.3 ((a5 b) →0 (a1 b)) = 1

Proof of Theorem lem3.3.3
StepHypRef Expression
1 df-i0 43 . 2 ((a5 b) →0 (a1 b)) = ((a5 b) ∪ (a1 b))
2 df-b1 1048 . . 3 (a1 b) = ((a1 b) ∩ (b1 a))
32lor 70 . 2 ((a5 b) ∪ (a1 b)) = ((a5 b) ∪ ((a1 b) ∩ (b1 a)))
4 lem3.3.3lem3 1051 . . 3 (a5 b) ≤ ((a1 b) ∩ (b1 a))
54sklem 230 . 2 ((a5 b) ∪ ((a1 b) ∩ (b1 a))) = 1
61, 3, 53tr 65 1 ((a5 b) →0 (a1 b)) = 1
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →0 wi0 11   →1 wi1 12   ≡5 wid5 22   ↔1 wb1 24 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-i0 43  df-i1 44  df-le1 130  df-le2 131  df-id5 1047  df-b1 1048 This theorem is referenced by:  lem3.3.5  1055
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