details

property	value
status	complete
benchmark	Ex16_Luc06_C.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n192.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.61355 seconds
cpu usage	3.46869
user time	3.30565
system time	0.163039
max virtual memory	1.8343604E7
max residence set size	225608.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSToCSRProof [EQUIVALENT, 0 ms] (2) CSR (3) CSRRRRProof [EQUIVALENT, 28 ms] (4) CSR (5) CSDependencyPairsProof [EQUIVALENT, 0 ms] (6) QCSDP (7) QCSDependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(X, X)) -> mark(f(a, b)) active(b) -> mark(a) active(f(X1, X2)) -> f(active(X1), X2) f(mark(X1), X2) -> mark(f(X1, X2)) proper(f(X1, X2)) -> f(proper(X1), proper(X2)) proper(a) -> ok(a) proper(b) -> ok(b) f(ok(X1), ok(X2)) -> ok(f(X1, X2)) 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(X, X)) -> mark(f(a, b)) active(b) -> mark(a) active(f(X1, X2)) -> f(active(X1), X2) f(mark(X1), X2) -> mark(f(X1, X2)) proper(f(X1, X2)) -> f(proper(X1), proper(X2)) proper(a) -> ok(a) proper(b) -> ok(b) f(ok(X1), ok(X2)) -> ok(f(X1, X2)) 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: {1} 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(X, X) -> f(a, b) b -> a The replacement map contains the following entries: f: {1} a: empty set b: empty set ---------------------------------------- (3) CSRRRRProof (EQUIVALENT) The following CSR is given: Context-sensitive rewrite system: The TRS R consists of the following rules: f(X, X) -> f(a, b) b -> a The replacement map contains the following entries: f: {1} a: empty set b: empty set Used ordering: Polynomial interpretation [POLO]: POL(a) = 0 POL(b) = 2 POL(f(x_1, x_2)) = x_1 popout

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all output return to TRS Standard