Theorem bdnthALT 17066

Description: Alternate proof of bdnth 17065 not using bdfal 17064. Then, bdfal 17064 can be proved from this theorem, using fal 1409. The total number of proof steps would be 17 (for bdnthALT 17066) + 3 = 20, which is more than 8 (for bdfal 17064) + 9 (for bdnth 17065) = 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 17063	. . 3 ⊢ BOUNDED ⊤
2	1	ax-bdn 17048	. 2 ⊢ BOUNDED ¬ ⊤
3		notnot 638	. . . 4 ⊢ (⊤ → ¬ ¬ ⊤)
4	3	mptru 1411	. . 3 ⊢ ¬ ¬ ⊤
5		bdnth.1	. . 3 ⊢ ¬ 𝜑
6	4, 5	2false 713	. 2 ⊢ (¬ ⊤ ↔ 𝜑)
7	2, 6	bd0 17055	1 ⊢ BOUNDED 𝜑

This proof depends on syntax axioms:  ¬ wn 3  ⊤wtru 1403  BOUNDED wbd 17043

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-bd0 17044  ax-bdim 17045  ax-bdn 17048  ax-bdeq 17051

This proof depends on definitions:  df-bi 117  df-tru 1405

This theorem is used by: (None)