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Theorem lem3.4.3 1076
Description: Equation 3.11 of [PavMeg1999] p. 9. (Contributed by Roy F. Longton, 29-Jun-2005.) (Revised by Roy F. Longton, 3-Jul-2005.)
Hypothesis
Ref Expression
lem3.4.3.1 (a2 b) = 1
Assertion
Ref Expression
lem3.4.3 (a2 (a5 b)) = 1

Proof of Theorem lem3.4.3
StepHypRef Expression
1 lem3.4.3.1 . . . . . 6 (a2 b) = 1
212vwomr2a 364 . . . . 5 (a1 b) = 1
32ax-r1 35 . . . 4 1 = (a1 b)
4 anidm 111 . . . . . . . . . 10 (aa) = a
54ax-r1 35 . . . . . . . . 9 a = (aa)
65ran 78 . . . . . . . 8 (ab) = ((aa) ∩ b)
7 lea 160 . . . . . . . . . 10 (aa) ≤ a
87lel 151 . . . . . . . . 9 ((aa) ∩ b) ≤ a
97leran 153 . . . . . . . . . 10 ((aa) ∩ b) ≤ (ab)
109ler 149 . . . . . . . . 9 ((aa) ∩ b) ≤ ((ab) ∪ (ab ))
118, 10ler2an 173 . . . . . . . 8 ((aa) ∩ b) ≤ (a ∩ ((ab) ∪ (ab )))
126, 11bltr 138 . . . . . . 7 (ab) ≤ (a ∩ ((ab) ∪ (ab )))
13 df-id5 1047 . . . . . . . . 9 (a5 b) = ((ab) ∪ (ab ))
1413ax-r1 35 . . . . . . . 8 ((ab) ∪ (ab )) = (a5 b)
1514lan 77 . . . . . . 7 (a ∩ ((ab) ∪ (ab ))) = (a ∩ (a5 b))
1612, 15lbtr 139 . . . . . 6 (ab) ≤ (a ∩ (a5 b))
1716lelor 166 . . . . 5 (a ∪ (ab)) ≤ (a ∪ (a ∩ (a5 b)))
18 df-i1 44 . . . . 5 (a1 b) = (a ∪ (ab))
19 df-i1 44 . . . . 5 (a1 (a5 b)) = (a ∪ (a ∩ (a5 b)))
2017, 18, 19le3tr1 140 . . . 4 (a1 b) ≤ (a1 (a5 b))
213, 20bltr 138 . . 3 1 ≤ (a1 (a5 b))
2221lem3.3.5lem 1054 . 2 (a1 (a5 b)) = 1
23222vwomr1a 363 1 (a2 (a5 b)) = 1
Colors of variables: term
Syntax hints:   = wb 1   wn 4  wo 6  wa 7  1wt 8  1 wi1 12  2 wi2 13  5 wid5 22
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  ax-wom 361
This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44  df-i2 45  df-le1 130  df-le2 131  df-id5 1047
This theorem is referenced by:  lem3.4.4  1077
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