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TRS Innermost pair #487093585
details
property
value
status
complete
benchmark
innermost5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n149.star.cs.uiowa.edu
space
Mixed_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.83462 seconds
cpu usage
4.11433
user time
3.94283
system time
0.171499
max virtual memory
1.8477524E7
max residence set size
261600.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) UsableRulesProof [EQUIVALENT, 0 ms] (21) QDP (22) QReductionProof [EQUIVALENT, 0 ms] (23) QDP (24) QDPQMonotonicMRRProof [EQUIVALENT, 25 ms] (25) QDP (26) UsableRulesProof [EQUIVALENT, 0 ms] (27) QDP (28) QDPOrderProof [EQUIVALENT, 0 ms] (29) QDP (30) PisEmptyProof [EQUIVALENT, 0 ms] (31) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(x), y, b) -> f(g(h(x)), y, i(y)) g(h(x)) -> g(x) g(s(x)) -> s(x) g(0) -> s(0) h(0) -> a i(0) -> b i(s(y)) -> i(y) The set Q consists of the following terms: f(s(x0), x1, b) g(h(x0)) g(s(x0)) g(0) h(0) i(0) i(s(x0)) ---------------------------------------- (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), y, b) -> F(g(h(x)), y, i(y)) F(s(x), y, b) -> G(h(x)) F(s(x), y, b) -> H(x) F(s(x), y, b) -> I(y) G(h(x)) -> G(x) I(s(y)) -> I(y) The TRS R consists of the following rules: f(s(x), y, b) -> f(g(h(x)), y, i(y)) g(h(x)) -> g(x) g(s(x)) -> s(x) g(0) -> s(0) h(0) -> a i(0) -> b i(s(y)) -> i(y) The set Q consists of the following terms: f(s(x0), x1, b) g(h(x0)) g(s(x0))
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