Theorem bdnthALT 15327

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.)

Hypothesis

Ref	Expression
bdnth.1	⊢ ¬ 𝜑

Assertion

Ref	Expression
bdnthALT	⊢ BOUNDED 𝜑

Proof of Theorem bdnthALT

Step	Hyp	Ref	Expression
1		bdtru 15324	. . 3 ⊢ BOUNDED ⊤
2	1	ax-bdn 15309	. 2 ⊢ BOUNDED ¬ ⊤
3		notnot 630	. . . 4 ⊢ (⊤ → ¬ ¬ ⊤)
4	3	mptru 1373	. . 3 ⊢ ¬ ¬ ⊤
5		bdnth.1	. . 3 ⊢ ¬ 𝜑
6	4, 5	2false 702	. 2 ⊢ (¬ ⊤ ↔ 𝜑)
7	2, 6	bd0 15316	1 ⊢ BOUNDED 𝜑

Syntax hints:  ¬ wn 3  ⊤wtru 1365  BOUNDED wbd 15304

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)