TRS Stand 20472 pair #381711455

details

property	value
status	complete
benchmark	IJCAR_26a.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n114.star.cs.uiowa.edu
space	AProVE_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.07824921608 seconds
cpu usage	5.147577741
max memory	3.5776512E8

stage attributes

key	value
output-size	19895
starexec-result	YES

output

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


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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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) MNOCProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) MRRProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) DependencyGraphProof [EQUIVALENT, 0 ms]
        (18) TRUE
    (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) QDPOrderProof [EQUIVALENT, 10 ms]
        (26) QDP
        (27) QDPOrderProof [EQUIVALENT, 8 ms]
        (28) QDP
        (29) TransformationProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) DependencyGraphProof [EQUIVALENT, 0 ms]
        (32) TRUE
    (33) QDP
        (34) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (35) YES


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

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

   p(0) -> 0
   p(s(x)) -> x
   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   plus(s(x), y) -> s(plus(p(s(x)), y))
   plus(x, s(y)) -> s(plus(x, p(s(y))))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PLUS(s(x), y) -> PLUS(x, y)
   PLUS(s(x), y) -> PLUS(p(s(x)), y)

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