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Mirrors > Home > ILE Home > Th. List > Mathboxes > bdnthALT | Unicode version |
Description: Alternate proof of bdnth 15326 not using bdfal 15325. Then, bdfal 15325 can be proved from this theorem, using fal 1371. The total number of proof steps would be 17 (for bdnthALT 15327) + 3 = 20, which is more than 8 (for bdfal 15325) + 9 (for bdnth 15326) = 17. (Contributed by BJ, 6-Oct-2019.) (Proof modification is discouraged.) (New usage is discouraged.) |
Ref | Expression |
---|---|
bdnth.1 |
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Ref | Expression |
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bdnthALT |
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Step | Hyp | Ref | Expression |
---|---|---|---|
1 | bdtru 15324 |
. . 3
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2 | 1 | ax-bdn 15309 |
. 2
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3 | notnot 630 |
. . . 4
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4 | 3 | mptru 1373 |
. . 3
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5 | bdnth.1 |
. . 3
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6 | 4, 5 | 2false 702 |
. 2
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7 | 2, 6 | bd0 15316 |
1
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Colors of variables: wff set class |
Syntax hints: ![]() ![]() |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-in1 615 ax-in2 616 ax-bd0 15305 ax-bdim 15306 ax-bdn 15309 ax-bdeq 15312 |
This theorem depends on definitions: df-bi 117 df-tru 1367 |
This theorem is referenced by: (None) |
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