details

property	value
status	complete
benchmark	2.26.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n076.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.25163507462 seconds
cpu usage	5.931879807
max memory	4.74189824E8

stage attributes

key	value
output-size	2162
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 103 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 1 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0) -> 0 f(s(0)) -> s(0) f(s(s(x))) -> p(h(g(x))) g(0) -> pair(s(0), s(0)) g(s(x)) -> h(g(x)) h(x) -> pair(+(p(x), q(x)), p(x)) p(pair(x, y)) -> x q(pair(x, y)) -> y +(x, 0) -> x +(x, s(y)) -> s(+(x, y)) f(s(s(x))) -> +(p(g(x)), q(g(x))) g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: f/1(YES) 0/0) s/1(YES) p/1)YES( h/1(YES) g/1(YES) pair/2(YES,YES) +/2(YES,YES) q/1)YES( Quasi precedence: f_1 > 0 > s_1 > pair_2 f_1 > g_1 > h_1 > +_2 > s_1 > pair_2 Status: f_1: [1] 0: multiset status s_1: multiset status h_1: multiset status g_1: multiset status pair_2: [2,1] +_2: multiset status With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(0) -> 0 f(s(0)) -> s(0) f(s(s(x))) -> p(h(g(x))) g(0) -> pair(s(0), s(0)) g(s(x)) -> h(g(x)) h(x) -> pair(+(p(x), q(x)), p(x)) p(pair(x, y)) -> x q(pair(x, y)) -> y +(x, 0) -> x +(x, s(y)) -> s(+(x, y)) f(s(s(x))) -> +(p(g(x)), q(g(x))) g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (3) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. popout

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all output return to TRS Standard