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TRS Stand 20472 pair #381709979
details
property
value
status
complete
benchmark
12.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n022.star.cs.uiowa.edu
space
Various_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.68582892418 seconds
cpu usage
3.932024041
max memory
2.405376E8
stage attributes
key
value
output-size
7042
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 28 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) MRRProof [EQUIVALENT, 0 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: O(0) -> 0 +(0, x) -> x +(x, 0) -> x +(O(x), O(y)) -> O(+(x, y)) +(O(x), I(y)) -> I(+(x, y)) +(I(x), O(y)) -> I(+(x, y)) +(I(x), I(y)) -> O(+(+(x, y), I(0))) *(0, x) -> 0 *(x, 0) -> 0 *(O(x), y) -> O(*(x, y)) *(I(x), y) -> +(O(*(x, y)), 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: +^1(O(x), O(y)) -> O^1(+(x, y)) +^1(O(x), O(y)) -> +^1(x, y) +^1(O(x), I(y)) -> +^1(x, y) +^1(I(x), O(y)) -> +^1(x, y) +^1(I(x), I(y)) -> O^1(+(+(x, y), I(0))) +^1(I(x), I(y)) -> +^1(+(x, y), I(0)) +^1(I(x), I(y)) -> +^1(x, y) *^1(O(x), y) -> O^1(*(x, y)) *^1(O(x), y) -> *^1(x, y) *^1(I(x), y) -> +^1(O(*(x, y)), y) *^1(I(x), y) -> O^1(*(x, y)) *^1(I(x), y) -> *^1(x, y) The TRS R consists of the following rules: O(0) -> 0 +(0, x) -> x +(x, 0) -> x +(O(x), O(y)) -> O(+(x, y)) +(O(x), I(y)) -> I(+(x, y)) +(I(x), O(y)) -> I(+(x, y)) +(I(x), I(y)) -> O(+(+(x, y), I(0))) *(0, x) -> 0 *(x, 0) -> 0 *(O(x), y) -> O(*(x, y)) *(I(x), y) -> +(O(*(x, y)), 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 2 SCCs with 5 less nodes. ---------------------------------------- (4)
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