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TRS Stand 20472 pair #381711449
details
property
value
status
complete
benchmark
thiemann26.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n074.star.cs.uiowa.edu
space
AProVE_07
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
61.9463200569 seconds
cpu usage
157.15579393
max memory
6.750236672E9
stage attributes
key
value
output-size
9194
starexec-result
NO
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) NonTerminationLoopProof [COMPLETE, 39.9 s] (17) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: ge(x, 0) -> true ge(0, s(y)) -> false ge(s(x), s(y)) -> ge(x, y) minus(x, 0) -> x minus(0, y) -> 0 minus(s(x), s(y)) -> minus(x, y) id_inc(x) -> x id_inc(x) -> s(x) quot(x, y) -> div(x, y, 0) div(x, y, z) -> if(ge(y, s(0)), ge(x, y), x, y, z) if(false, b, x, y, z) -> div_by_zero if(true, false, x, y, z) -> z if(true, true, x, y, z) -> div(minus(x, y), y, id_inc(z)) 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: GE(s(x), s(y)) -> GE(x, y) MINUS(s(x), s(y)) -> MINUS(x, y) QUOT(x, y) -> DIV(x, y, 0) DIV(x, y, z) -> IF(ge(y, s(0)), ge(x, y), x, y, z) DIV(x, y, z) -> GE(y, s(0)) DIV(x, y, z) -> GE(x, y) IF(true, true, x, y, z) -> DIV(minus(x, y), y, id_inc(z)) IF(true, true, x, y, z) -> MINUS(x, y) IF(true, true, x, y, z) -> ID_INC(z) The TRS R consists of the following rules: ge(x, 0) -> true ge(0, s(y)) -> false ge(s(x), s(y)) -> ge(x, y) minus(x, 0) -> x minus(0, y) -> 0 minus(s(x), s(y)) -> minus(x, y) id_inc(x) -> x id_inc(x) -> s(x) quot(x, y) -> div(x, y, 0) div(x, y, z) -> if(ge(y, s(0)), ge(x, y), x, y, z) if(false, b, x, y, z) -> div_by_zero if(true, false, x, y, z) -> z if(true, true, x, y, z) -> div(minus(x, y), y, id_inc(z)) 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 3 SCCs with 5 less nodes. ----------------------------------------
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