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TRS Stand 20472 pair #381715987
details
property
value
status
complete
benchmark
Ex9_Luc06_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n095.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.71953487396 seconds
cpu usage
3.709437083
max memory
2.3793664E8
stage attributes
key
value
output-size
4755
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSToCSRProof [EQUIVALENT, 0 ms] (2) CSR (3) CSDependencyPairsProof [EQUIVALENT, 0 ms] (4) QCSDP (5) QCSDependencyGraphProof [EQUIVALENT, 0 ms] (6) QCSDP (7) QCSUsableRulesProof [EQUIVALENT, 0 ms] (8) QCSDP (9) QCSDPInstantiationProcessor [EQUIVALENT, 0 ms] (10) QCSDP (11) PIsEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(a, X, X)) -> mark(f(X, b, b)) active(b) -> mark(a) active(f(X1, X2, X3)) -> f(X1, active(X2), X3) f(X1, mark(X2), X3) -> mark(f(X1, X2, X3)) proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3)) proper(a) -> ok(a) proper(b) -> ok(b) f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) Q is empty. ---------------------------------------- (1) QTRSToCSRProof (EQUIVALENT) The following Q TRS is given: Q restricted rewrite system: The TRS R consists of the following rules: active(f(a, X, X)) -> mark(f(X, b, b)) active(b) -> mark(a) active(f(X1, X2, X3)) -> f(X1, active(X2), X3) f(X1, mark(X2), X3) -> mark(f(X1, X2, X3)) proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3)) proper(a) -> ok(a) proper(b) -> ok(b) f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) Q is empty. Special symbols used for the transformation (see [GM04]): top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1 The replacement map contains the following entries: f: {2} a: empty set b: empty set The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound). ---------------------------------------- (2) Obligation: Context-sensitive rewrite system: The TRS R consists of the following rules: f(a, X, X) -> f(X, b, b) b -> a The replacement map contains the following entries: f: {2} a: empty set b: empty set ---------------------------------------- (3) CSDependencyPairsProof (EQUIVALENT) Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem. ---------------------------------------- (4) Obligation: Q-restricted context-sensitive dependency pair problem: For all symbols f in {f_3, F_3} we have mu(f) = {2}.
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