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Theorem ud1 595
 Description: Unified disjunction for Sasaki implication.
Assertion
Ref Expression
ud1 (ab) = ((a1 b) →1 (((a1 b) →1 (b1 a)) →1 a))

Proof of Theorem ud1
StepHypRef Expression
1 ud1lem1 560 . . . . . 6 ((a1 b) →1 (b1 a)) = (a ∪ (ab ))
21ud1lem0b 256 . . . . 5 (((a1 b) →1 (b1 a)) →1 a) = ((a ∪ (ab )) →1 a)
3 ud1lem2 561 . . . . 5 ((a ∪ (ab )) →1 a) = (ab)
42, 3ax-r2 36 . . . 4 (((a1 b) →1 (b1 a)) →1 a) = (ab)
54ud1lem0a 255 . . 3 ((a1 b) →1 (((a1 b) →1 (b1 a)) →1 a)) = ((a1 b) →1 (ab))
6 ud1lem3 562 . . 3 ((a1 b) →1 (ab)) = (ab)
75, 6ax-r2 36 . 2 ((a1 b) →1 (((a1 b) →1 (b1 a)) →1 a)) = (ab)
87ax-r1 35 1 (ab) = ((a1 b) →1 (((a1 b) →1 (b1 a)) →1 a))
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →1 wi1 12 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-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133 This theorem is referenced by: (None)
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