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TRS Innermost pair #487524655
details
property
value
status
complete
benchmark
test830.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
Mixed_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.68684482574 seconds
cpu usage
3.822006567
max memory
2.4383488E8
stage attributes
key
value
output-size
3972
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 66 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) DependencyPairsProof [EQUIVALENT, 16 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(X)) -> f(X) g(cons(0, Y)) -> g(Y) g(cons(s(X), Y)) -> s(X) h(cons(X, Y)) -> h(g(cons(X, Y))) The set Q consists of the following terms: f(s(x0)) g(cons(0, x0)) g(cons(s(x0), x1)) h(cons(x0, x1)) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 2 POL(cons(x_1, x_2)) = x_1 + x_2 POL(f(x_1)) = 2*x_1 POL(g(x_1)) = x_1 POL(h(x_1)) = 2*x_1 POL(s(x_1)) = 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: g(cons(0, Y)) -> g(Y) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(X)) -> f(X) g(cons(s(X), Y)) -> s(X) h(cons(X, Y)) -> h(g(cons(X, Y))) The set Q consists of the following terms: f(s(x0)) g(cons(0, x0)) g(cons(s(x0), x1)) h(cons(x0, x1)) ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(f(x_1)) = 2*x_1 POL(g(x_1)) = x_1 POL(h(x_1)) = x_1 POL(s(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(s(X)) -> f(X)
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