TRS Stand 20472 pair #381710636

details

property	value
status	complete
benchmark	JFP_Ex51.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n010.star.cs.uiowa.edu
space	AProVE_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.2493481636 seconds
cpu usage	5.794072394
max memory	3.81816832E8

stage attributes

key	value
output-size	14128
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) QDPOrderProof [EQUIVALENT, 75 ms]
        (17) QDP
        (18) DependencyGraphProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (21) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   minus_active(0, y) -> 0
   mark(0) -> 0
   minus_active(s(x), s(y)) -> minus_active(x, y)
   mark(s(x)) -> s(mark(x))
   ge_active(x, 0) -> true
   mark(minus(x, y)) -> minus_active(x, y)
   ge_active(0, s(y)) -> false
   mark(ge(x, y)) -> ge_active(x, y)
   ge_active(s(x), s(y)) -> ge_active(x, y)
   mark(div(x, y)) -> div_active(mark(x), y)
   div_active(0, s(y)) -> 0
   mark(if(x, y, z)) -> if_active(mark(x), y, z)
   div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
   if_active(true, x, y) -> mark(x)
   minus_active(x, y) -> minus(x, y)
   if_active(false, x, y) -> mark(y)
   ge_active(x, y) -> ge(x, y)
   if_active(x, y, z) -> if(x, y, z)
   div_active(x, y) -> div(x, y)

Q is empty.

----------------------------------------

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS_ACTIVE(s(x), s(y)) -> MINUS_ACTIVE(x, y)
   MARK(s(x)) -> MARK(x)
   MARK(minus(x, y)) -> MINUS_ACTIVE(x, y)
   MARK(ge(x, y)) -> GE_ACTIVE(x, y)
   GE_ACTIVE(s(x), s(y)) -> GE_ACTIVE(x, y)
   MARK(div(x, y)) -> DIV_ACTIVE(mark(x), y)
   MARK(div(x, y)) -> MARK(x)
   MARK(if(x, y, z)) -> IF_ACTIVE(mark(x), y, z)
   MARK(if(x, y, z)) -> MARK(x)
   DIV_ACTIVE(s(x), s(y)) -> IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
   DIV_ACTIVE(s(x), s(y)) -> GE_ACTIVE(x, y)
   IF_ACTIVE(true, x, y) -> MARK(x)
   IF_ACTIVE(false, x, y) -> MARK(y)

The TRS R consists of the following rules:

   minus_active(0, y) -> 0
   mark(0) -> 0
   minus_active(s(x), s(y)) -> minus_active(x, y)
   mark(s(x)) -> s(mark(x))
   ge_active(x, 0) -> true
   mark(minus(x, y)) -> minus_active(x, y)
   ge_active(0, s(y)) -> false

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