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TRS Inner 89993 pair #381734067
details
property
value
status
complete
benchmark
Ex1_2_Luc02c_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n112.star.cs.uiowa.edu
space
Transformed_CSR_innermost_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.03884792328 seconds
cpu usage
4.936057689
max memory
2.48274944E8
stage attributes
key
value
output-size
9967
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 115 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 15 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 0 ms] (10) QTRS (11) RisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(2nd(cons(X, cons(Y, Z)))) -> mark(Y) active(from(X)) -> mark(cons(X, from(s(X)))) active(2nd(X)) -> 2nd(active(X)) active(cons(X1, X2)) -> cons(active(X1), X2) active(from(X)) -> from(active(X)) active(s(X)) -> s(active(X)) 2nd(mark(X)) -> mark(2nd(X)) cons(mark(X1), X2) -> mark(cons(X1, X2)) from(mark(X)) -> mark(from(X)) s(mark(X)) -> mark(s(X)) proper(2nd(X)) -> 2nd(proper(X)) proper(cons(X1, X2)) -> cons(proper(X1), proper(X2)) proper(from(X)) -> from(proper(X)) proper(s(X)) -> s(proper(X)) 2nd(ok(X)) -> ok(2nd(X)) cons(ok(X1), ok(X2)) -> ok(cons(X1, X2)) from(ok(X)) -> ok(from(X)) s(ok(X)) -> ok(s(X)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) The set Q consists of the following terms: active(from(x0)) active(2nd(x0)) active(cons(x0, x1)) active(s(x0)) 2nd(mark(x0)) cons(mark(x0), x1) from(mark(x0)) s(mark(x0)) proper(2nd(x0)) proper(cons(x0, x1)) proper(from(x0)) proper(s(x0)) 2nd(ok(x0)) cons(ok(x0), ok(x1)) from(ok(x0)) s(ok(x0)) top(mark(x0)) top(ok(x0)) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(2nd(x_1)) = 2*x_1 POL(active(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(from(x_1)) = x_1 POL(mark(x_1)) = x_1 POL(ok(x_1)) = 1 + 2*x_1 POL(proper(x_1)) = x_1 POL(s(x_1)) = x_1 POL(top(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: 2nd(ok(X)) -> ok(2nd(X)) cons(ok(X1), ok(X2)) -> ok(cons(X1, X2)) top(ok(X)) -> top(active(X))
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