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TRS Innermost pair #516979115
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
n093.star.cs.uiowa.edu
space
Transformed_CSR_innermost_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.45311784744 seconds
cpu usage
7.007454291
max memory
4.98036736E8
stage attributes
key
value
output-size
7455
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPQMonotonicMRRProof [EQUIVALENT, 32 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 1 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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