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TRS Inner 89993 pair #381733502
details
property
value
status
complete
benchmark
#4.22.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n088.star.cs.uiowa.edu
space
AG01_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.63443589211 seconds
cpu usage
3.568983094
max memory
2.3216128E8
stage attributes
key
value
output-size
3462
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) QDP (5) QReductionProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPSizeChangeProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: quot(0, s(y), s(z)) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) quot(x, 0, s(z)) -> s(quot(x, s(z), s(z))) The set Q consists of the following terms: quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) quot(x0, 0, s(x1)) ---------------------------------------- (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: QUOT(s(x), s(y), z) -> QUOT(x, y, z) QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z)) The TRS R consists of the following rules: quot(0, s(y), s(z)) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) quot(x, 0, s(z)) -> s(quot(x, s(z), s(z))) The set Q consists of the following terms: quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) quot(x0, 0, s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: QUOT(s(x), s(y), z) -> QUOT(x, y, z) QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z)) R is empty. The set Q consists of the following terms: quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) quot(x0, 0, s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) quot(x0, 0, s(x1))
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