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Theorem ud5 599
 Description: Unified disjunction for relevance implication.
Assertion
Ref Expression
ud5 (ab) = ((a5 b) →5 (((a5 b) →5 (b5 a)) →5 a))

Proof of Theorem ud5
StepHypRef Expression
1 ud5lem1 589 . . . . . 6 ((a5 b) →5 (b5 a)) = (ab )
21ud5lem0b 265 . . . . 5 (((a5 b) →5 (b5 a)) →5 a) = ((ab ) →5 a)
3 ud5lem2 590 . . . . 5 ((ab ) →5 a) = (a ∪ (ab))
42, 3ax-r2 36 . . . 4 (((a5 b) →5 (b5 a)) →5 a) = (a ∪ (ab))
54ud5lem0a 264 . . 3 ((a5 b) →5 (((a5 b) →5 (b5 a)) →5 a)) = ((a5 b) →5 (a ∪ (ab)))
6 ud5lem3 594 . . 3 ((a5 b) →5 (a ∪ (ab))) = (ab)
75, 6ax-r2 36 . 2 ((a5 b) →5 (((a5 b) →5 (b5 a)) →5 a)) = (ab)
87ax-r1 35 1 (ab) = ((a5 b) →5 (((a5 b) →5 (b5 a)) →5 a))
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →5 wi5 16 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-i5 48  df-le1 130  df-le2 131  df-c1 132  df-c2 133 This theorem is referenced by: (None)
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