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TRS Inner 89993 pair #381733817
details
property
value
status
complete
benchmark
Ex4_4_Luc96b_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n092.star.cs.uiowa.edu
space
Transformed_CSR_innermost_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.1789329052 seconds
cpu usage
4.557697608
max memory
2.72150528E8
stage attributes
key
value
output-size
7459
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 28 ms] (2) QDP (3) QDPQMonotonicMRRProof [EQUIVALENT, 16 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) QDPOrderProof [EQUIVALENT, 0 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__f(g(X), Y) -> a__f(mark(X), f(g(X), Y)) mark(f(X1, X2)) -> a__f(mark(X1), X2) mark(g(X)) -> g(mark(X)) a__f(X1, X2) -> f(X1, X2) The set Q consists of the following terms: mark(f(x0, x1)) mark(g(x0)) a__f(x0, 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: A__F(g(X), Y) -> A__F(mark(X), f(g(X), Y)) A__F(g(X), Y) -> MARK(X) MARK(f(X1, X2)) -> A__F(mark(X1), X2) MARK(f(X1, X2)) -> MARK(X1) MARK(g(X)) -> MARK(X) The TRS R consists of the following rules: a__f(g(X), Y) -> a__f(mark(X), f(g(X), Y)) mark(f(X1, X2)) -> a__f(mark(X1), X2) mark(g(X)) -> g(mark(X)) a__f(X1, X2) -> f(X1, X2) The set Q consists of the following terms: mark(f(x0, x1)) mark(g(x0)) a__f(x0, x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPQMonotonicMRRProof (EQUIVALENT) By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain. Strictly oriented dependency pairs: MARK(f(X1, X2)) -> A__F(mark(X1), X2) MARK(f(X1, X2)) -> MARK(X1) Used ordering: Polynomial interpretation [POLO]: POL(A__F(x_1, x_2)) = 2*x_1 POL(MARK(x_1)) = 2*x_1 POL(a__f(x_1, x_2)) = 2 + 2*x_1
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