details

property	value
status	complete
benchmark	ex2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	AProVE_10

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	11.9248 seconds
cpu usage	37.9117
user time	36.3838
system time	1.52783
max virtual memory	1.9594784E7
max residence set size	3792340.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT,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, 25 ms] (18) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(tt, x) -> g(h(f(x, x)), s(x)) h(f(0, y)) -> tt f(s(x), y) -> f(x, s(y)) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is h(f(0, y)) -> tt f(s(x), y) -> f(x, s(y)) The TRS R 2 is g(tt, x) -> g(h(f(x, x)), s(x)) The signature Sigma is {g_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(tt, x) -> g(h(f(x, x)), s(x)) h(f(0, y)) -> tt f(s(x), y) -> f(x, s(y)) The set Q consists of the following terms: g(tt, x0) h(f(0, x0)) f(s(x0), x1) ---------------------------------------- (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: G(tt, x) -> G(h(f(x, x)), s(x)) G(tt, x) -> H(f(x, x)) G(tt, x) -> F(x, x) F(s(x), y) -> F(x, s(y)) The TRS R consists of the following rules: g(tt, x) -> g(h(f(x, x)), s(x)) h(f(0, y)) -> tt f(s(x), y) -> f(x, s(y)) The set Q consists of the following terms: g(tt, x0) h(f(0, x0)) f(s(x0), x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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all output return to TRS Standard