Theorem bdnthALT 16534

Description: Alternate proof of bdnth 16533 not using bdfal 16532. Then, bdfal 16532 can be proved from this theorem, using fal 1405. The total number of proof steps would be 17 (for bdnthALT 16534) + 3 = 20, which is more than 8 (for bdfal 16532) + 9 (for bdnth 16533) = 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 16531	. . 3 ⊢ BOUNDED ⊤
2	1	ax-bdn 16516	. 2 ⊢ BOUNDED ¬ ⊤
3		notnot 634	. . . 4 ⊢ (⊤ → ¬ ¬ ⊤)
4	3	mptru 1407	. . 3 ⊢ ¬ ¬ ⊤
5		bdnth.1	. . 3 ⊢ ¬ 𝜑
6	4, 5	2false 709	. 2 ⊢ (¬ ⊤ ↔ 𝜑)
7	2, 6	bd0 16523	1 ⊢ BOUNDED 𝜑

Syntax hints:  ¬ wn 3  ⊤wtru 1399  BOUNDED wbd 16511

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 619  ax-in2 620  ax-bd0 16512  ax-bdim 16513  ax-bdn 16516  ax-bdeq 16519

This theorem depends on definitions:  df-bi 117  df-tru 1401

This theorem is referenced by: (None)