Theorem bdnthALT 16128

Description: Alternate proof of bdnth 16127 not using bdfal 16126. Then, bdfal 16126 can be proved from this theorem, using fal 1402. The total number of proof steps would be 17 (for bdnthALT 16128) + 3 = 20, which is more than 8 (for bdfal 16126) + 9 (for bdnth 16127) = 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 16125	. . 3 ⊢ BOUNDED ⊤
2	1	ax-bdn 16110	. 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 16117	1 ⊢ BOUNDED 𝜑

Syntax hints:  ¬ wn 3  ⊤wtru 1396  BOUNDED wbd 16105

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 16106  ax-bdim 16107  ax-bdn 16110  ax-bdeq 16113

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

This theorem is referenced by: (None)