details

property	value
status	complete
benchmark	hom01.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n097.star.cs.uiowa.edu
space	Trafo_06

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	7.56186413765 seconds
cpu usage	26.779499823
max memory	1.43325184E9

stage attributes

key	value
output-size	6516
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 6 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 1 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 570 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 6 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(d(b(d(a(x1))))))) -> a(a(c(a(a(b(d(x1))))))) a(a(c(x1))) -> c(c(a(a(x1)))) c(c(c(x1))) -> b(d(c(b(d(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(d(b(d(b(a(a(x1))))))) -> d(b(a(a(c(a(a(x1))))))) c(a(a(x1))) -> a(a(c(c(x1)))) c(c(c(x1))) -> d(b(c(d(b(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(d(b(d(b(a(a(x1))))))) -> A(a(c(a(a(x1))))) A(d(b(d(b(a(a(x1))))))) -> A(c(a(a(x1)))) A(d(b(d(b(a(a(x1))))))) -> C(a(a(x1))) C(a(a(x1))) -> A(a(c(c(x1)))) C(a(a(x1))) -> A(c(c(x1))) C(a(a(x1))) -> C(c(x1)) C(a(a(x1))) -> C(x1) C(c(c(x1))) -> C(d(b(x1))) The TRS R consists of the following rules: a(d(b(d(b(a(a(x1))))))) -> d(b(a(a(c(a(a(x1))))))) c(a(a(x1))) -> a(a(c(c(x1)))) c(c(c(x1))) -> d(b(c(d(b(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 1 less node. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(d(b(d(b(a(a(x1))))))) -> A(c(a(a(x1)))) popout

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