Theorem bdnthALT 14557

Description: Alternate proof of bdnth 14556 not using bdfal 14555. Then, bdfal 14555 can be proved from this theorem, using fal 1360. The total number of proof steps would be 17 (for bdnthALT 14557) + 3 = 20, which is more than 8 (for bdfal 14555) + 9 (for bdnth 14556) = 17. (Contributed by BJ, 6-Oct-2019.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
bdnth.1	|- -. ph

Assertion

Ref	Expression
bdnthALT	|- BOUNDED ph

Proof of Theorem bdnthALT

Step	Hyp	Ref	Expression
1		bdtru 14554	. . 3 |- BOUNDED T.
2	1	ax-bdn 14539	. 2 |- BOUNDED -. T.
3		notnot 629	. . . 4 |- ( T. -> -. -. T. )
4	3	mptru 1362	. . 3 |- -. -. T.
5		bdnth.1	. . 3 |- -. ph
6	4, 5	2false 701	. 2 |- ( -. T. <-> ph )
7	2, 6	bd0 14546	1 |- BOUNDED ph

Syntax hints:   -. wn 3   T. wtru 1354  BOUNDED wbd 14534

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 614  ax-in2 615  ax-bd0 14535  ax-bdim 14536  ax-bdn 14539  ax-bdeq 14542

This theorem depends on definitions:  df-bi 117  df-tru 1356

This theorem is referenced by: (None)