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TRS Inner 89993 pair #381733656
details
property
value
status
complete
benchmark
#4.34.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n108.star.cs.uiowa.edu
space
AG01_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.72077202797 seconds
cpu usage
3.651800623
max memory
2.38194688E8
stage attributes
key
value
output-size
6682
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 1 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 0 ms] (14) QDP (15) PisEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0) -> true f(1) -> false f(s(x)) -> f(x) if(true, x, y) -> x if(false, x, y) -> y g(s(x), s(y)) -> if(f(x), s(x), s(y)) g(x, c(y)) -> g(x, g(s(c(y)), y)) The set Q consists of the following terms: f(0) f(1) f(s(x0)) if(true, x0, x1) if(false, x0, x1) g(s(x0), s(x1)) g(x0, c(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: F(s(x)) -> F(x) G(s(x), s(y)) -> IF(f(x), s(x), s(y)) G(s(x), s(y)) -> F(x) G(x, c(y)) -> G(x, g(s(c(y)), y)) G(x, c(y)) -> G(s(c(y)), y) The TRS R consists of the following rules: f(0) -> true f(1) -> false f(s(x)) -> f(x) if(true, x, y) -> x if(false, x, y) -> y g(s(x), s(y)) -> if(f(x), s(x), s(y)) g(x, c(y)) -> g(x, g(s(c(y)), y)) The set Q consists of the following terms: f(0) f(1) f(s(x0)) if(true, x0, x1) if(false, x0, x1) g(s(x0), s(x1)) g(x0, c(x1)) 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 2 less nodes.
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