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Theorem bdnthALT 13022
 Description: Alternate proof of bdnth 13021 not using bdfal 13020. Then, bdfal 13020 can be proved from this theorem, using fal 1338. The total number of proof steps would be 17 (for bdnthALT 13022) + 3 = 20, which is more than 8 (for bdfal 13020) + 9 (for bdnth 13021) = 17. (Contributed by BJ, 6-Oct-2019.) (Proof modification is discouraged.) (New usage is discouraged.)
Hypothesis
Ref Expression
bdnth.1
Assertion
Ref Expression
bdnthALT BOUNDED

Proof of Theorem bdnthALT
StepHypRef Expression
1 bdtru 13019 . . 3 BOUNDED
21ax-bdn 13004 . 2 BOUNDED
3 notnot 618 . . . 4
43mptru 1340 . . 3
5 bdnth.1 . . 3
64, 52false 690 . 2
72, 6bd0 13011 1 BOUNDED
 Colors of variables: wff set class Syntax hints:   wn 3   wtru 1332  BOUNDED wbd 12999 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-bd0 13000  ax-bdim 13001  ax-bdn 13004  ax-bdeq 13007 This theorem depends on definitions:  df-bi 116  df-tru 1334 This theorem is referenced by: (None)
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