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Theorem lem4.6.6i1j3 1094
 Description: Equation 4.14 of [MegPav2000] p. 23. The variable i in the paper is set to 1, and j is set to 3. (Contributed by Roy F. Longton, 1-Jul-2005.)
Assertion
Ref Expression
lem4.6.6i1j3 ((a1 b) ∪ (a3 b)) = (a0 b)

Proof of Theorem lem4.6.6i1j3
StepHypRef Expression
1 leo 158 . . . . . . . 8 a ≤ (a ∪ (ab))
21ler 149 . . . . . . 7 a ≤ ((a ∪ (ab)) ∪ ((ab) ∪ (ab )))
32lecom 180 . . . . . 6 a C ((a ∪ (ab)) ∪ ((ab) ∪ (ab )))
43comcom6 459 . . . . 5 a C ((a ∪ (ab)) ∪ ((ab) ∪ (ab )))
54comcom 453 . . . 4 ((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) C a
6 lear 161 . . . . . . 7 (ab) ≤ b
76lelor 166 . . . . . 6 (a ∪ (ab)) ≤ (ab)
8 lea 160 . . . . . . . 8 (ab) ≤ a
9 lea 160 . . . . . . . 8 (ab ) ≤ a
108, 9lel2or 170 . . . . . . 7 ((ab) ∪ (ab )) ≤ a
1110ler 149 . . . . . 6 ((ab) ∪ (ab )) ≤ (ab)
127, 11lel2or 170 . . . . 5 ((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ≤ (ab)
1312lecom 180 . . . 4 ((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) C (ab)
145, 13fh3 471 . . 3 (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (a ∩ (ab))) = ((((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ a) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab)))
15 ax-a3 32 . . 3 (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (a ∩ (ab))) = ((a ∪ (ab)) ∪ (((ab) ∪ (ab )) ∪ (a ∩ (ab))))
16 ax-a2 31 . . . . 5 (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ a) = (a ∪ ((a ∪ (ab)) ∪ ((ab) ∪ (ab ))))
1716ran 78 . . . 4 ((((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ a) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab))) = ((a ∪ ((a ∪ (ab)) ∪ ((ab) ∪ (ab )))) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab)))
18 ax-a3 32 . . . . . 6 ((a ∪ (a ∪ (ab))) ∪ ((ab) ∪ (ab ))) = (a ∪ ((a ∪ (ab)) ∪ ((ab) ∪ (ab ))))
1918ax-r1 35 . . . . 5 (a ∪ ((a ∪ (ab)) ∪ ((ab) ∪ (ab )))) = ((a ∪ (a ∪ (ab))) ∪ ((ab) ∪ (ab )))
2019ran 78 . . . 4 ((a ∪ ((a ∪ (ab)) ∪ ((ab) ∪ (ab )))) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab))) = (((a ∪ (a ∪ (ab))) ∪ ((ab) ∪ (ab ))) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab)))
21 ax-a3 32 . . . . . . . 8 ((aa ) ∪ (ab)) = (a ∪ (a ∪ (ab)))
2221ax-r1 35 . . . . . . 7 (a ∪ (a ∪ (ab))) = ((aa ) ∪ (ab))
2322ax-r5 38 . . . . . 6 ((a ∪ (a ∪ (ab))) ∪ ((ab) ∪ (ab ))) = (((aa ) ∪ (ab)) ∪ ((ab) ∪ (ab )))
2423ran 78 . . . . 5 (((a ∪ (a ∪ (ab))) ∪ ((ab) ∪ (ab ))) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab))) = ((((aa ) ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab)))
25 ax-a4 33 . . . . . . . . . 10 (1 ∪ (aa )) = (aa )
2625df-le1 130 . . . . . . . . 9 1 ≤ (aa )
2726ler 149 . . . . . . . 8 1 ≤ ((aa ) ∪ (ab))
2827ler 149 . . . . . . 7 1 ≤ (((aa ) ∪ (ab)) ∪ ((ab) ∪ (ab )))
2928lem3.3.5lem 1054 . . . . . 6 (((aa ) ∪ (ab)) ∪ ((ab) ∪ (ab ))) = 1
3029ran 78 . . . . 5 ((((aa ) ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab))) = (1 ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab)))
31 an1r 107 . . . . . 6 (1 ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab))) = (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab))
32 ax-a2 31 . . . . . . 7 ((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) = (((ab) ∪ (ab )) ∪ (a ∪ (ab)))
3332ax-r5 38 . . . . . 6 (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab)) = ((((ab) ∪ (ab )) ∪ (a ∪ (ab))) ∪ (ab))
34 ax-a3 32 . . . . . . 7 ((((ab) ∪ (ab )) ∪ (a ∪ (ab))) ∪ (ab)) = (((ab) ∪ (ab )) ∪ ((a ∪ (ab)) ∪ (ab)))
35 orordi 112 . . . . . . . . . 10 (a ∪ ((ab) ∪ b)) = ((a ∪ (ab)) ∪ (ab))
3635ax-r1 35 . . . . . . . . 9 ((a ∪ (ab)) ∪ (ab)) = (a ∪ ((ab) ∪ b))
3736lor 70 . . . . . . . 8 (((ab) ∪ (ab )) ∪ ((a ∪ (ab)) ∪ (ab))) = (((ab) ∪ (ab )) ∪ (a ∪ ((ab) ∪ b)))
386df-le2 131 . . . . . . . . . 10 ((ab) ∪ b) = b
3938lor 70 . . . . . . . . 9 (a ∪ ((ab) ∪ b)) = (ab)
4039lor 70 . . . . . . . 8 (((ab) ∪ (ab )) ∪ (a ∪ ((ab) ∪ b))) = (((ab) ∪ (ab )) ∪ (ab))
4111df-le2 131 . . . . . . . 8 (((ab) ∪ (ab )) ∪ (ab)) = (ab)
4237, 40, 413tr 65 . . . . . . 7 (((ab) ∪ (ab )) ∪ ((a ∪ (ab)) ∪ (ab))) = (ab)
4334, 42ax-r2 36 . . . . . 6 ((((ab) ∪ (ab )) ∪ (a ∪ (ab))) ∪ (ab)) = (ab)
4431, 33, 433tr 65 . . . . 5 (1 ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab))) = (ab)
4524, 30, 443tr 65 . . . 4 (((a ∪ (a ∪ (ab))) ∪ ((ab) ∪ (ab ))) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab))) = (ab)
4617, 20, 453tr 65 . . 3 ((((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ a) ∩ (((a ∪ (ab)) ∪ ((ab) ∪ (ab ))) ∪ (ab))) = (ab)
4714, 15, 463tr2 64 . 2 ((a ∪ (ab)) ∪ (((ab) ∪ (ab )) ∪ (a ∩ (ab)))) = (ab)
48 df-i1 44 . . 3 (a1 b) = (a ∪ (ab))
49 df-i3 46 . . 3 (a3 b) = (((ab) ∪ (ab )) ∪ (a ∩ (ab)))
5048, 492or 72 . 2 ((a1 b) ∪ (a3 b)) = ((a ∪ (ab)) ∪ (((ab) ∪ (ab )) ∪ (a ∩ (ab))))
51 df-i0 43 . 2 (a0 b) = (ab)
5247, 50, 513tr1 63 1 ((a1 b) ∪ (a3 b)) = (a0 b)
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →0 wi0 11   →1 wi1 12   →3 wi3 14 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-r3 439 This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i0 43  df-i1 44  df-i3 46  df-le1 130  df-le2 131  df-c1 132  df-c2 133 This theorem is referenced by: (None)
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