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SRS Standard pair #516975940
details
property
value
status
complete
benchmark
beans3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n010.star.cs.uiowa.edu
space
Zantema_06
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
5.50902318954 seconds
cpu usage
18.740116146
max memory
1.265283072E9
stage attributes
key
value
output-size
12279
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 16 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 2 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 11 ms] (7) QDP (8) QDPOrderProof [EQUIVALENT, 19 ms] (9) QDP (10) DependencyGraphProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPOrderProof [EQUIVALENT, 48 ms] (13) QDP (14) QDPOrderProof [EQUIVALENT, 37 ms] (15) QDP (16) DependencyGraphProof [EQUIVALENT, 0 ms] (17) TRUE (18) QDP (19) UsableRulesProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPOrderProof [EQUIVALENT, 38 ms] (22) QDP (23) PisEmptyProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) b(c(a(x1))) -> a(b(c(x1))) c(b(x1)) -> d(x1) a(d(x1)) -> d(a(x1)) d(x1) -> b(a(x1)) L(a(a(x1))) -> L(a(b(c(x1)))) c(R(x1)) -> c(b(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: B(a(a(x1))) -> A(b(c(x1))) B(a(a(x1))) -> B(c(x1)) B(a(a(x1))) -> C(x1) C(a(x1)) -> A(c(x1)) C(a(x1)) -> C(x1) B(c(a(x1))) -> A(b(c(x1))) B(c(a(x1))) -> B(c(x1)) B(c(a(x1))) -> C(x1) C(b(x1)) -> D(x1) A(d(x1)) -> D(a(x1)) A(d(x1)) -> A(x1) D(x1) -> B(a(x1)) D(x1) -> A(x1) L^1(a(a(x1))) -> L^1(a(b(c(x1)))) L^1(a(a(x1))) -> A(b(c(x1))) L^1(a(a(x1))) -> B(c(x1)) L^1(a(a(x1))) -> C(x1) C(R(x1)) -> C(b(R(x1))) C(R(x1)) -> B(R(x1)) The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) b(c(a(x1))) -> a(b(c(x1))) c(b(x1)) -> d(x1) a(d(x1)) -> d(a(x1)) d(x1) -> b(a(x1)) L(a(a(x1))) -> L(a(b(c(x1)))) c(R(x1)) -> c(b(R(x1)))
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