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Theorem ska2a 226
Description: Axiom KA2a in Pavicic and Megill, 1998
Assertion
Ref Expression
ska2a (((a v c) == (b v c)) == ((c v a) == (c v b))) = 1
Proof of Theorem ska2a
StepHypRef Expression
1 ax-a2 31 . . 3 (a v c) = (c v a)
2 ax-a2 31 . . 3 (b v c) = (c v b)
31, 22bi 99 . 2 ((a v c) == (b v c)) = ((c v a) == (c v b))
43bi1 118 1 (((a v c) == (b v c)) == ((c v a) == (c v b))) = 1
Colors of variables: term
Syntax hints:   = wb 1   == tb 5   v wo 6  1wt 8
This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38
This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42
This theorem is referenced by: (None)
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