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Runtime_Complexity_Innermost_Rewriting 2019-04-01 06.40 pair #433313549
details
property
value
status
complete
benchmark
22.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n107.star.cs.uiowa.edu
space
Various_04
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
6.626 seconds
cpu usage
20.1837
user time
18.7274
system time
1.45632
max virtual memory
1.9008948E7
max residence set size
4233860.0
stage attributes
key
value
starexec-result
WORST_CASE(Omega(n^1), O(n^1))
output
19.06/5.74 WORST_CASE(Omega(n^1), O(n^1)) 19.06/5.75 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 19.06/5.75 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 19.06/5.75 19.06/5.75 19.06/5.75 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1). 19.06/5.75 19.06/5.75 (0) CpxTRS 19.06/5.75 (1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms] 19.06/5.75 (2) CdtProblem 19.06/5.75 (3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms] 19.06/5.75 (4) CdtProblem 19.06/5.75 (5) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms] 19.06/5.75 (6) CdtProblem 19.06/5.75 (7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 43 ms] 19.06/5.75 (8) CdtProblem 19.06/5.75 (9) CdtNarrowingProof [BOTH BOUNDS(ID, ID), 0 ms] 19.06/5.75 (10) CdtProblem 19.06/5.75 (11) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms] 19.06/5.75 (12) CdtProblem 19.06/5.75 (13) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms] 19.06/5.75 (14) CdtProblem 19.06/5.75 (15) CdtLeafRemovalProof [ComplexityIfPolyImplication, 0 ms] 19.06/5.75 (16) CdtProblem 19.06/5.75 (17) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms] 19.06/5.75 (18) CdtProblem 19.06/5.75 (19) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 3 ms] 19.06/5.75 (20) CdtProblem 19.06/5.75 (21) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms] 19.06/5.75 (22) BOUNDS(1, 1) 19.06/5.75 (23) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] 19.06/5.75 (24) TRS for Loop Detection 19.06/5.75 (25) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] 19.06/5.75 (26) BEST 19.06/5.75 (27) proven lower bound 19.06/5.75 (28) LowerBoundPropagationProof [FINISHED, 0 ms] 19.06/5.75 (29) BOUNDS(n^1, INF) 19.06/5.75 (30) TRS for Loop Detection 19.06/5.75 19.06/5.75 19.06/5.75 ---------------------------------------- 19.06/5.75 19.06/5.75 (0) 19.06/5.75 Obligation: 19.06/5.75 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1). 19.06/5.75 19.06/5.75 19.06/5.75 The TRS R consists of the following rules: 19.06/5.75 19.06/5.75 f(x, 0) -> s(0) 19.06/5.75 f(s(x), s(y)) -> s(f(x, y)) 19.06/5.75 g(0, x) -> g(f(x, x), x) 19.06/5.75 19.06/5.75 S is empty. 19.06/5.75 Rewrite Strategy: INNERMOST 19.06/5.75 ---------------------------------------- 19.06/5.75 19.06/5.75 (1) CpxTrsToCdtProof (UPPER BOUND(ID)) 19.06/5.75 Converted Cpx (relative) TRS to CDT 19.06/5.75 ---------------------------------------- 19.06/5.75 19.06/5.75 (2) 19.06/5.75 Obligation: 19.06/5.75 Complexity Dependency Tuples Problem 19.06/5.75 19.06/5.75 Rules: 19.06/5.75 f(z0, 0) -> s(0) 19.06/5.75 f(s(z0), s(z1)) -> s(f(z0, z1)) 19.06/5.75 g(0, z0) -> g(f(z0, z0), z0) 19.06/5.75 Tuples: 19.06/5.75 F(z0, 0) -> c 19.06/5.75 F(s(z0), s(z1)) -> c1(F(z0, z1)) 19.06/5.75 G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0)) 19.06/5.75 S tuples: 19.06/5.75 F(z0, 0) -> c 19.06/5.75 F(s(z0), s(z1)) -> c1(F(z0, z1)) 19.06/5.75 G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0)) 19.06/5.75 K tuples:none 19.06/5.75 Defined Rule Symbols: f_2, g_2 19.06/5.75 19.06/5.75 Defined Pair Symbols: F_2, G_2 19.06/5.75 19.06/5.75 Compound Symbols: c, c1_1, c2_2 19.06/5.75 19.06/5.75 19.06/5.75 ---------------------------------------- 19.06/5.75 19.06/5.75 (3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID)) 19.06/5.75 Removed 1 trailing nodes: 19.06/5.75 F(z0, 0) -> c 19.06/5.75 19.06/5.75 ---------------------------------------- 19.06/5.75 19.06/5.75 (4) 19.06/5.75 Obligation: 19.06/5.75 Complexity Dependency Tuples Problem 19.06/5.75 19.06/5.75 Rules: 19.06/5.75 f(z0, 0) -> s(0) 19.06/5.75 f(s(z0), s(z1)) -> s(f(z0, z1))
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