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Theorem sac 835
Description: Theorem showing "Sasaki complement" is an operation.
Hypothesis
Ref Expression
sac.1 (a ->1 c) = (b ->1 c)
Assertion
Ref Expression
sac (a' ->1 c) = (b' ->1 c)
Proof of Theorem sac
StepHypRef Expression
1 sac.1 . . 3 (a ->1 c) = (b ->1 c)
21ud1lem0b 256 . 2 ((a ->1 c) ->1 c) = ((b ->1 c) ->1 c)
3 u1lem12 781 . 2 ((a ->1 c) ->1 c) = (a' ->1 c)
4 u1lem12 781 . 2 ((b ->1 c) ->1 c) = (b' ->1 c)
52, 3, 43tr2 64 1 (a' ->1 c) = (b' ->1 c)
Colors of variables: term
Syntax hints:   = wb 1  'wn 4   ->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:  negantlem9  859  negant3  860  negant4  862
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