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TRS Stand 20472 pair #381716517
details
property
value
status
complete
benchmark
Ex5_Zan97_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n020.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.56739616394 seconds
cpu usage
3.561170755
max memory
2.2833152E8
stage attributes
key
value
output-size
2362
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 0 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4)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: f(X) -> if(X, c, n__f(true)) if(true, X, Y) -> X if(false, X, Y) -> activate(Y) f(X) -> n__f(X) activate(n__f(X)) -> f(X) activate(X) -> X Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(activate(x_1)) = 2 + 2*x_1 POL(c) = 0 POL(f(x_1)) = 1 + 2*x_1 POL(false) = 2 POL(if(x_1, x_2, x_3)) = 1 + 2*x_1 + x_2 + 2*x_3 POL(n__f(x_1)) = x_1 POL(true) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: if(true, X, Y) -> X if(false, X, Y) -> activate(Y) f(X) -> n__f(X) activate(n__f(X)) -> f(X) activate(X) -> X ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(X) -> if(X, c, n__f(true)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:true > n__f_1 > c > f_1 > if_3 and weight map: c=1 true=1 f_1=3 n__f_1=1 if_3=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: f(X) -> if(X, c, n__f(true)) ---------------------------------------- (4) Obligation: Q restricted rewrite system: R is empty.
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