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Theorem ud2lem3 565
 Description: Lemma for unified disjunction.
Assertion
Ref Expression
ud2lem3 ((a2 b) →2 (ab)) = (ab)

Proof of Theorem ud2lem3
StepHypRef Expression
1 df-i2 45 . 2 ((a2 b) →2 (ab)) = ((ab) ∪ ((a2 b) ∩ (ab) ))
2 ud2lem0c 278 . . . . 5 (a2 b) = (b ∩ (ab))
32ran 78 . . . 4 ((a2 b) ∩ (ab) ) = ((b ∩ (ab)) ∩ (ab) )
43lor 70 . . 3 ((ab) ∪ ((a2 b) ∩ (ab) )) = ((ab) ∪ ((b ∩ (ab)) ∩ (ab) ))
5 coman2 186 . . . . . 6 (b ∩ (ab)) C (ab)
65comcom 453 . . . . 5 (ab) C (b ∩ (ab))
7 comid 187 . . . . . 6 (ab) C (ab)
87comcom2 183 . . . . 5 (ab) C (ab)
96, 8fh3 471 . . . 4 ((ab) ∪ ((b ∩ (ab)) ∩ (ab) )) = (((ab) ∪ (b ∩ (ab))) ∩ ((ab) ∪ (ab) ))
10 ancom 74 . . . . . . 7 (b ∩ (ab)) = ((ab) ∩ b )
1110lor 70 . . . . . 6 ((ab) ∪ (b ∩ (ab))) = ((ab) ∪ ((ab) ∩ b ))
12 df-t 41 . . . . . . 7 1 = ((ab) ∪ (ab) )
1312ax-r1 35 . . . . . 6 ((ab) ∪ (ab) ) = 1
1411, 132an 79 . . . . 5 (((ab) ∪ (b ∩ (ab))) ∩ ((ab) ∪ (ab) )) = (((ab) ∪ ((ab) ∩ b )) ∩ 1)
15 an1 106 . . . . . 6 (((ab) ∪ ((ab) ∩ b )) ∩ 1) = ((ab) ∪ ((ab) ∩ b ))
16 orabs 120 . . . . . 6 ((ab) ∪ ((ab) ∩ b )) = (ab)
1715, 16ax-r2 36 . . . . 5 (((ab) ∪ ((ab) ∩ b )) ∩ 1) = (ab)
1814, 17ax-r2 36 . . . 4 (((ab) ∪ (b ∩ (ab))) ∩ ((ab) ∪ (ab) )) = (ab)
199, 18ax-r2 36 . . 3 ((ab) ∪ ((b ∩ (ab)) ∩ (ab) )) = (ab)
204, 19ax-r2 36 . 2 ((ab) ∪ ((a2 b) ∩ (ab) )) = (ab)
211, 20ax-r2 36 1 ((a2 b) →2 (ab)) = (ab)
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →2 wi2 13 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-i2 45  df-le1 130  df-le2 131  df-c1 132  df-c2 133 This theorem is referenced by:  ud2  596
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