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Theorem u1lemob 630
 Description: Lemma for Sasaki implication study. (Contributed by NM, 15-Dec-1997.)
Assertion
Ref Expression
u1lemob ((a1 b) ∪ b) = (ab)

Proof of Theorem u1lemob
StepHypRef Expression
1 df-i1 44 . . 3 (a1 b) = (a ∪ (ab))
21ax-r5 38 . 2 ((a1 b) ∪ b) = ((a ∪ (ab)) ∪ b)
3 or32 82 . . 3 ((a ∪ (ab)) ∪ b) = ((ab) ∪ (ab))
4 ax-a2 31 . . . 4 ((ab) ∪ (ab)) = ((ab) ∪ (ab))
5 lear 161 . . . . . 6 (ab) ≤ b
6 leor 159 . . . . . 6 b ≤ (ab)
75, 6letr 137 . . . . 5 (ab) ≤ (ab)
87df-le2 131 . . . 4 ((ab) ∪ (ab)) = (ab)
94, 8ax-r2 36 . . 3 ((ab) ∪ (ab)) = (ab)
103, 9ax-r2 36 . 2 ((a ∪ (ab)) ∪ b) = (ab)
112, 10ax-r2 36 1 ((a1 b) ∪ b) = (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-a 40  df-i1 44  df-le1 130  df-le2 131 This theorem is referenced by:  u1lemnanb  655  u12lem  771  salem1  837
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