TRS Standard pair #487067228

details

property	value
status	complete
benchmark	emmes-nonloop-ex1_1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n186.star.cs.uiowa.edu
space	EEG_IJCAR_12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	12.0485 seconds
cpu usage	38.4558
user time	37.0823
system time	1.37344
max virtual memory	1.8865556E7
max residence set size	3874028.0

stage attributes

key	value
starexec-result	NO

output

NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) MNOCProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) NonLoopProof [COMPLETE, 128 ms]
        (18) NO


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

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

   f(true, x, y) -> f(gt(x, y), plus2(x), plus1(y))
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)
   plus1(x) -> s(x)
   plus2(x) -> s(s(x))

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)
   plus1(x) -> s(x)
   plus2(x) -> s(s(x))

The TRS R 2 is 
   f(true, x, y) -> f(gt(x, y), plus2(x), plus1(y))

The signature Sigma is {f_3}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(true, x, y) -> f(gt(x, y), plus2(x), plus1(y))
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)
   plus1(x) -> s(x)
   plus2(x) -> s(s(x))

The set Q consists of the following terms:

   f(true, x0, x1)
   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   plus1(x0)
   plus2(x0)


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

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

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(true, x, y) -> F(gt(x, y), plus2(x), plus1(y))
   F(true, x, y) -> GT(x, y)
   F(true, x, y) -> PLUS2(x)
   F(true, x, y) -> PLUS1(y)
   GT(s(x), s(y)) -> GT(x, y)

The TRS R consists of the following rules:

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