details

property	value
status	complete
benchmark	size-12-alpha-3-num-16.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.21466 seconds
cpu usage	9.67
user time	9.22721
system time	0.442796
max virtual memory	4.0433448E7
max residence set size	1169264.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 129 ms] (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) b(x1) -> x1 b(a(c(x1))) -> c(c(a(a(b(x1))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) b(x1) -> x1 c(a(b(x1))) -> b(a(a(c(c(x1))))) Q is empty. ---------------------------------------- (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: A(x1) -> B(x1) C(a(b(x1))) -> B(a(a(c(c(x1))))) C(a(b(x1))) -> A(a(c(c(x1)))) C(a(b(x1))) -> A(c(c(x1))) C(a(b(x1))) -> C(c(x1)) C(a(b(x1))) -> C(x1) The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) b(x1) -> x1 c(a(b(x1))) -> b(a(a(c(c(x1))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: C(a(b(x1))) -> C(x1) C(a(b(x1))) -> C(c(x1)) The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) b(x1) -> x1 popout

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