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Theorem lem3.3.7i0e1 1057
Description: Equation 3.7 of [PavMeg1999] p. 9. The variable i in the paper is set to 0, and this is the first part of the equation. (Contributed by Roy F. Longton, 3-Jul-05.)
Assertion
Ref Expression
lem3.3.7i0e1 (a0 (ab)) = (a0 (ab))

Proof of Theorem lem3.3.7i0e1
StepHypRef Expression
1 or1 104 . . . . . . 7 (b ∪ 1) = 1
21ax-r1 35 . . . . . 6 1 = (b ∪ 1)
32lan 77 . . . . 5 ((a ∪ (ab)) ∩ 1) = ((a ∪ (ab)) ∩ (b ∪ 1))
4 an1 106 . . . . 5 ((a ∪ (ab)) ∩ 1) = (a ∪ (ab))
5 df-t 41 . . . . . . 7 1 = (aa )
65lor 70 . . . . . 6 (b ∪ 1) = (b ∪ (aa ))
76lan 77 . . . . 5 ((a ∪ (ab)) ∩ (b ∪ 1)) = ((a ∪ (ab)) ∩ (b ∪ (aa )))
83, 4, 73tr2 64 . . . 4 (a ∪ (ab)) = ((a ∪ (ab)) ∩ (b ∪ (aa )))
9 ax-a2 31 . . . . . 6 (aa ) = (aa)
109lor 70 . . . . 5 (b ∪ (aa )) = (b ∪ (aa))
1110lan 77 . . . 4 ((a ∪ (ab)) ∩ (b ∪ (aa ))) = ((a ∪ (ab)) ∩ (b ∪ (aa)))
12 ax-a3 32 . . . . . 6 ((ba ) ∪ a) = (b ∪ (aa))
1312ax-r1 35 . . . . 5 (b ∪ (aa)) = ((ba ) ∪ a)
1413lan 77 . . . 4 ((a ∪ (ab)) ∩ (b ∪ (aa))) = ((a ∪ (ab)) ∩ ((ba ) ∪ a))
158, 11, 143tr 65 . . 3 (a ∪ (ab)) = ((a ∪ (ab)) ∩ ((ba ) ∪ a))
16 ax-a2 31 . . . . 5 (ba ) = (ab )
1716ax-r5 38 . . . 4 ((ba ) ∪ a) = ((ab ) ∪ a)
1817lan 77 . . 3 ((a ∪ (ab)) ∩ ((ba ) ∪ a)) = ((a ∪ (ab)) ∩ ((ab ) ∪ a))
19 oran3 93 . . . . 5 (ab ) = (ab)
2019ax-r5 38 . . . 4 ((ab ) ∪ a) = ((ab)a)
2120lan 77 . . 3 ((a ∪ (ab)) ∩ ((ab ) ∪ a)) = ((a ∪ (ab)) ∩ ((ab)a))
2215, 18, 213tr 65 . 2 (a ∪ (ab)) = ((a ∪ (ab)) ∩ ((ab)a))
23 df-i0 43 . 2 (a0 (ab)) = (a ∪ (ab))
24 df-id0 49 . 2 (a0 (ab)) = ((a ∪ (ab)) ∩ ((ab)a))
2522, 23, 243tr1 63 1 (a0 (ab)) = (a0 (ab))
Colors of variables: term
Syntax hints:   = wb 1   wn 4  wo 6  wa 7  1wt 8  0 wi0 11  0 wid0 17
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-id0 49
This theorem is referenced by: (None)
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