TRS Stand 20472 pair #381712964

details

property	value
status	complete
benchmark	wiehe01.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n101.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.57392811775 seconds
cpu usage	6.509306363
max memory	3.36539648E8

stage attributes

key	value
output-size	12269
starexec-result	YES

output

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


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


YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QDPOrderProof [EQUIVALENT, 77 ms]
        (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 40 ms]
        (14) QDP
        (15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (16) YES
    (17) QDP
        (18) UsableRulesProof [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:

   times(x, 0) -> 0
   times(x, s(y)) -> plus(times(x, y), x)
   plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
   plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x))
   plus(zero, y) -> y
   plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y))))
   id(x) -> x
   if(true, x, y) -> x
   if(false, x, y) -> y
   not(x) -> if(x, false, true)
   gt(s(x), zero) -> true
   gt(zero, y) -> false
   gt(s(x), s(y)) -> gt(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:

   TIMES(x, s(y)) -> PLUS(times(x, y), x)
   TIMES(x, s(y)) -> TIMES(x, y)
   PLUS(s(x), s(y)) -> PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
   PLUS(s(x), s(y)) -> IF(gt(x, y), x, y)
   PLUS(s(x), s(y)) -> GT(x, y)
   PLUS(s(x), s(y)) -> IF(not(gt(x, y)), id(x), id(y))
   PLUS(s(x), s(y)) -> NOT(gt(x, y))
   PLUS(s(x), s(y)) -> ID(x)
   PLUS(s(x), s(y)) -> ID(y)
   PLUS(s(x), x) -> PLUS(if(gt(x, x), id(x), id(x)), s(x))
   PLUS(s(x), x) -> IF(gt(x, x), id(x), id(x))
   PLUS(s(x), x) -> GT(x, x)
   PLUS(s(x), x) -> ID(x)
   PLUS(id(x), s(y)) -> PLUS(x, if(gt(s(y), y), y, s(y)))
   PLUS(id(x), s(y)) -> IF(gt(s(y), y), y, s(y))
   PLUS(id(x), s(y)) -> GT(s(y), y)
   NOT(x) -> IF(x, false, true)
   GT(s(x), s(y)) -> GT(x, y)

The TRS R consists of the following rules:

   times(x, 0) -> 0
   times(x, s(y)) -> plus(times(x, y), x)
   plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
   plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x))
   plus(zero, y) -> y
   plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y))))
   id(x) -> x
   if(true, x, y) -> x

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