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Mirrors > Home > ILE Home > Th. List > Mathboxes > bdnthALT | Unicode version |
Description: Alternate proof of bdnth 9954 not using bdfal 9953. Then, bdfal 9953 can be proved from this theorem, using fal 1250. The total number of proof steps would be 17 (for bdnthALT 9955) + 3 = 20, which is more than 8 (for bdfal 9953) + 9 (for bdnth 9954) = 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 |
---|---|
bdnthALT |
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Step | Hyp | Ref | Expression |
---|---|---|---|
1 | bdtru 9952 |
. . 3
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2 | 1 | ax-bdn 9937 |
. 2
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3 | notnot 559 |
. . . 4
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4 | 3 | trud 1252 |
. . 3
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5 | bdnth.1 |
. . 3
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6 | 4, 5 | 2false 617 |
. 2
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7 | 2, 6 | bd0 9944 |
1
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Colors of variables: wff set class |
Syntax hints: ![]() ![]() |
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 9933 ax-bdim 9934 ax-bdn 9937 ax-bdeq 9940 |
This theorem depends on definitions: df-bi 110 df-tru 1246 |
This theorem is referenced by: (None) |
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