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Theorem lem3.3.7i2e2 1064
Description: Equation 3.7 of [PavMeg1999] p. 9. The variable i in the paper is set to 2, and this is the second part of the equation. (Contributed by Roy F. Longton, 3-Jul-2005.)
Assertion
Ref Expression
lem3.3.7i2e2 (a2 (ab)) = ((ab) ≡2 a)

Proof of Theorem lem3.3.7i2e2
StepHypRef Expression
1 oran3 93 . . . . . 6 (ab ) = (ab)
21ax-r1 35 . . . . 5 (ab) = (ab )
32lor 70 . . . 4 (a ∪ (ab) ) = (a ∪ (ab ))
43ran 78 . . 3 ((a ∪ (ab) ) ∩ ((ab) ∪ (a ∩ (ab) ))) = ((a ∪ (ab )) ∩ ((ab) ∪ (a ∩ (ab) )))
5 ax-a3 32 . . . . 5 ((aa ) ∪ b ) = (a ∪ (ab ))
65ax-r1 35 . . . 4 (a ∪ (ab )) = ((aa ) ∪ b )
76ran 78 . . 3 ((a ∪ (ab )) ∩ ((ab) ∪ (a ∩ (ab) ))) = (((aa ) ∪ b ) ∩ ((ab) ∪ (a ∩ (ab) )))
8 df-t 41 . . . . . . 7 1 = (aa )
98ax-r1 35 . . . . . 6 (aa ) = 1
109ax-r5 38 . . . . 5 ((aa ) ∪ b ) = (1 ∪ b )
1110ran 78 . . . 4 (((aa ) ∪ b ) ∩ ((ab) ∪ (a ∩ (ab) ))) = ((1 ∪ b ) ∩ ((ab) ∪ (a ∩ (ab) )))
12 or1r 105 . . . . 5 (1 ∪ b ) = 1
1312ran 78 . . . 4 ((1 ∪ b ) ∩ ((ab) ∪ (a ∩ (ab) ))) = (1 ∩ ((ab) ∪ (a ∩ (ab) )))
14 an1r 107 . . . . 5 (1 ∩ ((ab) ∪ (a ∩ (ab) ))) = ((ab) ∪ (a ∩ (ab) ))
15 anor3 90 . . . . . 6 (a ∩ (ab) ) = (a ∪ (ab))
1615lor 70 . . . . 5 ((ab) ∪ (a ∩ (ab) )) = ((ab) ∪ (a ∪ (ab)) )
17 orabs 120 . . . . . . . 8 (a ∪ (ab)) = a
1817ax-r4 37 . . . . . . 7 (a ∪ (ab)) = a
1918lor 70 . . . . . 6 ((ab) ∪ (a ∪ (ab)) ) = ((ab) ∪ a )
20 an1 106 . . . . . . . . 9 (((ab) ∪ a ) ∩ 1) = ((ab) ∪ a )
2120ax-r1 35 . . . . . . . 8 ((ab) ∪ a ) = (((ab) ∪ a ) ∩ 1)
228lan 77 . . . . . . . 8 (((ab) ∪ a ) ∩ 1) = (((ab) ∪ a ) ∩ (aa ))
2321, 22ax-r2 36 . . . . . . 7 ((ab) ∪ a ) = (((ab) ∪ a ) ∩ (aa ))
24 lea 160 . . . . . . . . . . . 12 (ab) ≤ a
2524df-le2 131 . . . . . . . . . . 11 ((ab) ∪ a) = a
2625ax-r1 35 . . . . . . . . . 10 a = ((ab) ∪ a)
2726ax-r4 37 . . . . . . . . 9 a = ((ab) ∪ a)
2827lor 70 . . . . . . . 8 (aa ) = (a ∪ ((ab) ∪ a) )
2928lan 77 . . . . . . 7 (((ab) ∪ a ) ∩ (aa )) = (((ab) ∪ a ) ∩ (a ∪ ((ab) ∪ a) ))
30 anor3 90 . . . . . . . . . 10 ((ab)a ) = ((ab) ∪ a)
3130ax-r1 35 . . . . . . . . 9 ((ab) ∪ a) = ((ab)a )
3231lor 70 . . . . . . . 8 (a ∪ ((ab) ∪ a) ) = (a ∪ ((ab)a ))
3332lan 77 . . . . . . 7 (((ab) ∪ a ) ∩ (a ∪ ((ab) ∪ a) )) = (((ab) ∪ a ) ∩ (a ∪ ((ab)a )))
3423, 29, 333tr 65 . . . . . 6 ((ab) ∪ a ) = (((ab) ∪ a ) ∩ (a ∪ ((ab)a )))
3519, 34ax-r2 36 . . . . 5 ((ab) ∪ (a ∪ (ab)) ) = (((ab) ∪ a ) ∩ (a ∪ ((ab)a )))
3614, 16, 353tr 65 . . . 4 (1 ∩ ((ab) ∪ (a ∩ (ab) ))) = (((ab) ∪ a ) ∩ (a ∪ ((ab)a )))
3711, 13, 363tr 65 . . 3 (((aa ) ∪ b ) ∩ ((ab) ∪ (a ∩ (ab) ))) = (((ab) ∪ a ) ∩ (a ∪ ((ab)a )))
384, 7, 373tr 65 . 2 ((a ∪ (ab) ) ∩ ((ab) ∪ (a ∩ (ab) ))) = (((ab) ∪ a ) ∩ (a ∪ ((ab)a )))
39 df-id2 51 . 2 (a2 (ab)) = ((a ∪ (ab) ) ∩ ((ab) ∪ (a ∩ (ab) )))
40 df-id2 51 . 2 ((ab) ≡2 a) = (((ab) ∪ a ) ∩ (a ∪ ((ab)a )))
4138, 39, 403tr1 63 1 (a2 (ab)) = ((ab) ≡2 a)
Colors of variables: term
Syntax hints:   = wb 1   wn 4  wo 6  wa 7  1wt 8  2 wid2 19
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-id2 51  df-le1 130  df-le2 131
This theorem is referenced by: (None)
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