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Theorem ud2 596
 Description: Unified disjunction for Dishkant implication.
Assertion
Ref Expression
ud2 (ab) = ((a2 b) →2 (((a2 b) →2 (b2 a)) →2 a))

Proof of Theorem ud2
StepHypRef Expression
1 ud2lem1 563 . . . . . 6 ((a2 b) →2 (b2 a)) = (a ∪ (ab ))
21ud2lem0b 259 . . . . 5 (((a2 b) →2 (b2 a)) →2 a) = ((a ∪ (ab )) →2 a)
3 ud2lem2 564 . . . . 5 ((a ∪ (ab )) →2 a) = (ab)
42, 3ax-r2 36 . . . 4 (((a2 b) →2 (b2 a)) →2 a) = (ab)
54ud2lem0a 258 . . 3 ((a2 b) →2 (((a2 b) →2 (b2 a)) →2 a)) = ((a2 b) →2 (ab))
6 ud2lem3 565 . . 3 ((a2 b) →2 (ab)) = (ab)
75, 6ax-r2 36 . 2 ((a2 b) →2 (((a2 b) →2 (b2 a)) →2 a)) = (ab)
87ax-r1 35 1 (ab) = ((a2 b) →2 (((a2 b) →2 (b2 a)) →2 a))
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →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: (None)
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