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Theorem bdnthALT 9270
Description: Alternate proof of bdnth 9269 not using bdfal 9268. Then, bdfal 9268 can be proved from this theorem, using fal 1249. The total number of proof steps would be 17 (for bdnthALT 9270) + 3 = 20, which is more than 8 (for bdfal 9268) + 9 (for bdnth 9269) = 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 9267 . . 3 BOUNDED
21ax-bdn 9252 . 2 BOUNDED
3 notnot1 559 . . . 4
43trud 1251 . . 3
5 bdnth.1 . . 3
64, 52false 616 . 2
72, 6bd0 9259 1 BOUNDED
Colors of variables: wff set class
Syntax hints:   wn 3   wtru 1243  BOUNDED wbd 9247
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-bd0 9248  ax-bdim 9249  ax-bdn 9252  ax-bdeq 9255
This theorem depends on definitions:  df-bi 110  df-tru 1245
This theorem is referenced by: (None)
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