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Theorem u1lemona 625
 Description: Lemma for Sasaki implication study. (Contributed by NM, 15-Dec-1997.)
Assertion
Ref Expression
u1lemona ((a1 b) ∪ a ) = (a ∪ (ab))

Proof of Theorem u1lemona
StepHypRef Expression
1 df-i1 44 . . 3 (a1 b) = (a ∪ (ab))
21ax-r5 38 . 2 ((a1 b) ∪ a ) = ((a ∪ (ab)) ∪ a )
3 or32 82 . . 3 ((a ∪ (ab)) ∪ a ) = ((aa ) ∪ (ab))
4 oridm 110 . . . 4 (aa ) = a
54ax-r5 38 . . 3 ((aa ) ∪ (ab)) = (a ∪ (ab))
63, 5ax-r2 36 . 2 ((a ∪ (ab)) ∪ a ) = (a ∪ (ab))
72, 6ax-r2 36 1 ((a1 b) ∪ a ) = (a ∪ (ab))
 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-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38 This theorem depends on definitions:  df-t 41  df-f 42  df-i1 44 This theorem is referenced by:  u1lemnaa  640  u1lem4  757
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