details

property	value
status	complete
benchmark	thiemann40_modified.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n174.star.cs.uiowa.edu
space	AProVE_08

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	13.3414 seconds
cpu usage	27.9267
user time	26.3391
system time	1.58759
max virtual memory	3.6846596E7
max residence set size	4367024.0

stage attributes

key	value
starexec-result	YES

output

YES 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 1 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPOrderProof [EQUIVALENT, 9 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: nonZero(0) -> false nonZero(s(x)) -> true p(s(0)) -> 0 p(s(s(x))) -> s(p(s(x))) id_inc(x) -> x id_inc(x) -> s(x) random(x) -> rand(x, 0) rand(x, y) -> if(nonZero(x), x, y) if(false, x, y) -> y if(true, x, y) -> rand(p(x), id_inc(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: P(s(s(x))) -> P(s(x)) RANDOM(x) -> RAND(x, 0) RAND(x, y) -> IF(nonZero(x), x, y) RAND(x, y) -> NONZERO(x) IF(true, x, y) -> RAND(p(x), id_inc(y)) IF(true, x, y) -> P(x) IF(true, x, y) -> ID_INC(y) The TRS R consists of the following rules: nonZero(0) -> false nonZero(s(x)) -> true p(s(0)) -> 0 p(s(s(x))) -> s(p(s(x))) id_inc(x) -> x id_inc(x) -> s(x) random(x) -> rand(x, 0) rand(x, y) -> if(nonZero(x), x, y) if(false, x, y) -> y if(true, x, y) -> rand(p(x), id_inc(y)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: P(s(s(x))) -> P(s(x)) The TRS R consists of the following rules: nonZero(0) -> false nonZero(s(x)) -> true popout

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all output return to TRS Standard