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Theorem cancel 892
 Description: Cancellation law eliminating →1 consequent.
Hypothesis
Ref Expression
cancel.1 ((d ∪ (a1 c)) →1 c) = ((d ∪ (b1 c)) →1 c)
Assertion
Ref Expression
cancel (d ∪ (a1 c)) = (d ∪ (b1 c))

Proof of Theorem cancel
StepHypRef Expression
1 cancel.1 . . 3 ((d ∪ (a1 c)) →1 c) = ((d ∪ (b1 c)) →1 c)
21cancellem 891 . 2 (d ∪ (a1 c)) ≤ (d ∪ (b1 c))
31ax-r1 35 . . 3 ((d ∪ (b1 c)) →1 c) = ((d ∪ (a1 c)) →1 c)
43cancellem 891 . 2 (d ∪ (b1 c)) ≤ (d ∪ (a1 c))
52, 4lebi 145 1 (d ∪ (a1 c)) = (d ∪ (b1 c))
 Colors of variables: term Syntax hints:   = wb 1   ∪ wo 6   →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: (None)
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