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TRS Innermost pair #487524561
details
property
value
status
complete
benchmark
Ex6_Luc98_L.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n072.star.cs.uiowa.edu
space
Transformed_CSR_innermost_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.66218900681 seconds
cpu usage
3.566652337
max memory
2.28216832E8
stage attributes
key
value
output-size
2453
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 60 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) RisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: first(0, X) -> nil first(s(X), cons(Y)) -> cons(Y) from(X) -> cons(X) The set Q consists of the following terms: first(0, x0) first(s(x0), cons(x1)) from(x0) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(cons(x_1)) = 2 + 2*x_1 POL(first(x_1, x_2)) = x_1 + 2*x_2 POL(from(x_1)) = 2 + 2*x_1 POL(nil) = 0 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: first(s(X), cons(Y)) -> cons(Y) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: first(0, X) -> nil from(X) -> cons(X) The set Q consists of the following terms: first(0, x0) first(s(x0), cons(x1)) from(x0) ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:first_2 > from_1 > cons_1 > nil > 0 and weight map: 0=1 nil=2 from_1=1 cons_1=1 first_2=0 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: first(0, X) -> nil from(X) -> cons(X) ----------------------------------------
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