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TRS Standard pair #516960980
details
property
value
status
complete
benchmark
thiemann40.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n064.star.cs.uiowa.edu
space
AProVE_07
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.42195796967 seconds
cpu usage
6.424094428
max memory
4.76131328E8
stage attributes
key
value
output-size
4205
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 53 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: nonZero(0) -> false nonZero(s(x)) -> true p(0) -> 0 p(s(x)) -> x id_inc(x) -> x id_inc(x) -> s(x) random(x) -> rand(x, 0) rand(x, y) -> if(nonZero(x), x, y) if(false, x, y) -> y if(true, x, y) -> rand(p(x), id_inc(y)) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: RANDOM(x) -> RAND(x, 0) RAND(x, y) -> IF(nonZero(x), x, y) RAND(x, y) -> NONZERO(x) IF(true, x, y) -> RAND(p(x), id_inc(y)) IF(true, x, y) -> P(x) IF(true, x, y) -> ID_INC(y) The TRS R consists of the following rules: nonZero(0) -> false nonZero(s(x)) -> true p(0) -> 0 p(s(x)) -> x id_inc(x) -> x id_inc(x) -> s(x) random(x) -> rand(x, 0) rand(x, y) -> if(nonZero(x), x, y) if(false, x, y) -> y if(true, x, y) -> rand(p(x), id_inc(y)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x, y) -> RAND(p(x), id_inc(y)) RAND(x, y) -> IF(nonZero(x), x, y) The TRS R consists of the following rules: nonZero(0) -> false nonZero(s(x)) -> true p(0) -> 0 p(s(x)) -> x id_inc(x) -> x id_inc(x) -> s(x) random(x) -> rand(x, 0)
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