Theorem bdnthALT 16340

Description: Alternate proof of bdnth 16339 not using bdfal 16338. Then, bdfal 16338 can be proved from this theorem, using fal 1402. The total number of proof steps would be 17 (for bdnthALT 16340) + 3 = 20, which is more than 8 (for bdfal 16338) + 9 (for bdnth 16339) = 17. (Contributed by BJ, 6-Oct-2019.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
bdnth.1	⊢ ¬ 𝜑

Assertion

Ref	Expression
bdnthALT	⊢ BOUNDED 𝜑

Proof of Theorem bdnthALT

Step	Hyp	Ref	Expression
1		bdtru 16337	. . 3 ⊢ BOUNDED ⊤
2	1	ax-bdn 16322	. 2 ⊢ BOUNDED ¬ ⊤
3		notnot 632	. . . 4 ⊢ (⊤ → ¬ ¬ ⊤)
4	3	mptru 1404	. . 3 ⊢ ¬ ¬ ⊤
5		bdnth.1	. . 3 ⊢ ¬ 𝜑
6	4, 5	2false 706	. 2 ⊢ (¬ ⊤ ↔ 𝜑)
7	2, 6	bd0 16329	1 ⊢ BOUNDED 𝜑

Syntax hints:  ¬ wn 3  ⊤wtru 1396  BOUNDED wbd 16317

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 617  ax-in2 618  ax-bd0 16318  ax-bdim 16319  ax-bdn 16322  ax-bdeq 16325

This theorem depends on definitions:  df-bi 117  df-tru 1398

This theorem is referenced by: (None)