Theorem bdnthALT 16959

Description: Alternate proof of bdnth 16958 not using bdfal 16957. Then, bdfal 16957 can be proved from this theorem, using fal 1409. The total number of proof steps would be 17 (for bdnthALT 16959) + 3 = 20, which is more than 8 (for bdfal 16957) + 9 (for bdnth 16958) = 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 16956	. . 3 ⊢ BOUNDED ⊤
2	1	ax-bdn 16941	. 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 16948	1 ⊢ BOUNDED 𝜑

This proof depends on syntax axioms:  ¬ wn 3  ⊤wtru 1403  BOUNDED wbd 16936

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 16937  ax-bdim 16938  ax-bdn 16941  ax-bdeq 16944

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

This theorem is used by: (None)