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TRS Standard pair #487068198
details
property
value
status
complete
benchmark
2.39.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n178.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.95386 seconds
cpu usage
4.65342
user time
4.4091
system time
0.244312
max virtual memory
1.8541468E7
max residence set size
299328.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 56 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 7 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 0 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 7 ms] (10) QTRS (11) QTRSRRRProof [EQUIVALENT, 0 ms] (12) QTRS (13) QTRSRRRProof [EQUIVALENT, 0 ms] (14) QTRS (15) QTRSRRRProof [EQUIVALENT, 0 ms] (16) QTRS (17) RisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: rev(nil) -> nil rev(.(x, y)) -> ++(rev(y), .(x, nil)) car(.(x, y)) -> x cdr(.(x, y)) -> y null(nil) -> true null(.(x, y)) -> false ++(nil, y) -> y ++(.(x, y), z) -> .(x, ++(y, z)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(++(x_1, x_2)) = x_1 + x_2 POL(.(x_1, x_2)) = 2*x_1 + x_2 POL(car(x_1)) = 2*x_1 POL(cdr(x_1)) = 2*x_1 POL(false) = 0 POL(nil) = 0 POL(null(x_1)) = 1 + 2*x_1 POL(rev(x_1)) = 2*x_1 POL(true) = 1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: null(.(x, y)) -> false ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: rev(nil) -> nil rev(.(x, y)) -> ++(rev(y), .(x, nil)) car(.(x, y)) -> x cdr(.(x, y)) -> y null(nil) -> true ++(nil, y) -> y ++(.(x, y), z) -> .(x, ++(y, z)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(++(x_1, x_2)) = x_1 + x_2 POL(.(x_1, x_2)) = 2*x_1 + x_2 POL(car(x_1)) = 2*x_1 POL(cdr(x_1)) = 1 + 2*x_1 POL(nil) = 0 POL(null(x_1)) = 2 + 2*x_1 POL(rev(x_1)) = 2*x_1 POL(true) = 2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
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