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Theorem u1lemnana 645
Description: Lemma for Sasaki implication study. (Contributed by NM, 15-Dec-1997.)
Assertion
Ref Expression
u1lemnana ((a1 b)a ) = 0

Proof of Theorem u1lemnana
StepHypRef Expression
1 anor3 90 . . 3 ((a1 b)a ) = ((a1 b) ∪ a)
2 u1lemoa 620 . . . 4 ((a1 b) ∪ a) = 1
32ax-r4 37 . . 3 ((a1 b) ∪ a) = 1
41, 3ax-r2 36 . 2 ((a1 b)a ) = 1
5 df-f 42 . . 3 0 = 1
65ax-r1 35 . 2 1 = 0
74, 6ax-r2 36 1 ((a1 b)a ) = 0
Colors of variables: term
Syntax hints:   = wb 1   wn 4  wo 6  wa 7  1wt 8  0wf 9  1 wi1 12
This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38
This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44
This theorem is referenced by: (None)
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