details

property	value
status	complete
benchmark	12.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n192.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	67.8459219933 seconds
cpu usage	267.662308133
max memory	8.652288E9

stage attributes

key	value
output-size	8278
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 35 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 48 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 9 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 408 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(l(x1)) -> l(a(x1)) a(c(x1)) -> c(a(x1)) c(a(r(x1))) -> r(a(x1)) l(r(a(a(x1)))) -> a(a(l(c(c(c(r(x1))))))) 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: A(l(x1)) -> L(a(x1)) A(l(x1)) -> A(x1) A(c(x1)) -> C(a(x1)) A(c(x1)) -> A(x1) C(a(r(x1))) -> A(x1) L(r(a(a(x1)))) -> A(a(l(c(c(c(r(x1))))))) L(r(a(a(x1)))) -> A(l(c(c(c(r(x1)))))) L(r(a(a(x1)))) -> L(c(c(c(r(x1))))) L(r(a(a(x1)))) -> C(c(c(r(x1)))) L(r(a(a(x1)))) -> C(c(r(x1))) L(r(a(a(x1)))) -> C(r(x1)) The TRS R consists of the following rules: a(l(x1)) -> l(a(x1)) a(c(x1)) -> c(a(x1)) c(a(r(x1))) -> r(a(x1)) l(r(a(a(x1)))) -> a(a(l(c(c(c(r(x1))))))) 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 1 SCC with 4 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: L(r(a(a(x1)))) -> A(a(l(c(c(c(r(x1))))))) A(l(x1)) -> L(a(x1)) L(r(a(a(x1)))) -> A(l(c(c(c(r(x1)))))) A(l(x1)) -> A(x1) A(c(x1)) -> C(a(x1)) C(a(r(x1))) -> A(x1) A(c(x1)) -> A(x1) The TRS R consists of the following rules: a(l(x1)) -> l(a(x1)) popout

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