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TRS Inner 89993 pair #381733491
details
property
value
status
complete
benchmark
test10.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n109.star.cs.uiowa.edu
space
Mixed_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.74185705185 seconds
cpu usage
3.79201782
max memory
1.96390912E8
stage attributes
key
value
output-size
4635
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 65 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 16 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: h(X, Z) -> f(X, s(X), Z) f(X, Y, g(X, Y)) -> h(0, g(X, Y)) g(0, Y) -> 0 g(X, s(Y)) -> g(X, Y) The set Q consists of the following terms: h(x0, x1) f(x0, x1, g(x0, x1)) g(0, x0) g(x0, s(x1)) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(f(x_1, x_2, x_3)) = x_1 + x_2 + x_3 POL(g(x_1, x_2)) = 2 + 2*x_1 + x_2 POL(h(x_1, x_2)) = 2*x_1 + x_2 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: g(0, Y) -> 0 ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: h(X, Z) -> f(X, s(X), Z) f(X, Y, g(X, Y)) -> h(0, g(X, Y)) g(X, s(Y)) -> g(X, Y) The set Q consists of the following terms: h(x0, x1) f(x0, x1, g(x0, x1)) g(0, x0) g(x0, s(x1)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: H(X, Z) -> F(X, s(X), Z) F(X, Y, g(X, Y)) -> H(0, g(X, Y)) G(X, s(Y)) -> G(X, Y)
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