Runtime Complexity: TRS Innermost pair #487112526

details

property	value
status	complete
benchmark	#4.32.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	5.63662 seconds
cpu usage	18.1802
user time	16.7722
system time	1.40799
max virtual memory	1.880942E7
max residence set size	4029328.0

stage attributes

key	value
starexec-result	WORST_CASE(Omega(n^1), O(n^1))

output

WORST_CASE(Omega(n^1), O(n^1))
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 7 ms]
(2) CdtProblem
    (3) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 54 ms]
    (4) CdtProblem
    (5) CdtInstantiationProof [BOTH BOUNDS(ID, ID), 0 ms]
    (6) CdtProblem
    (7) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms]
    (8) CdtProblem
    (9) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
    (10) CdtProblem
    (11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 0 ms]
    (12) CdtProblem
    (13) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
    (14) BOUNDS(1, 1)
(15) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(16) TRS for Loop Detection
    (17) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
    (18) BEST
        (19) proven lower bound
            (20) LowerBoundPropagationProof [FINISHED, 0 ms]
            (21) BOUNDS(n^1, INF)
        (22) TRS for Loop Detection


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(0)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   f(x, c(y)) -> f(x, s(f(y, y)))
   f(s(x), s(y)) -> f(x, s(c(s(y))))

S is empty.
Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS to CDT
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(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, c(z1)) -> f(z0, s(f(z1, z1)))
   f(s(z0), s(z1)) -> f(z0, s(c(s(z1))))
Tuples:
   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
   F(s(z0), s(z1)) -> c2(F(z0, s(c(s(z1)))))
S tuples:
   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
   F(s(z0), s(z1)) -> c2(F(z0, s(c(s(z1)))))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c1_2, c2_1


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(3) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
We considered the (Usable) Rules:none
And the Tuples:
   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
   F(s(z0), s(z1)) -> c2(F(z0, s(c(s(z1)))))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(F(x_1, x_2)) = x_2
   POL(c(x_1)) = [1] + x_1
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c2(x_1)) = x_1
   POL(f(x_1, x_2)) = [1] + x_2
   POL(s(x_1)) = 0

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:

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