YES(O(1), O(n^1)) 0.00/1.01 YES(O(1), O(n^1)) 0.00/1.04 0.00/1.04 0.00/1.04 0.00/1.04 0.00/1.04 0.00/1.04 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/1.04 0.00/1.04 0.00/1.04
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
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↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
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↳2 CdtProblem
0.00/1.04 
↳3 CdtUnreachableProof (⇔)
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↳4 CdtProblem
0.00/1.04 
↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
0.00/1.04 
↳6 CdtProblem
0.00/1.04 
↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
0.00/1.04 
↳8 CdtProblem
0.00/1.04 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/1.04 
↳10 CdtProblem
0.00/1.04 
↳11 CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
0.00/1.04 
↳12 CdtProblem
0.00/1.04 
↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/1.04 
↳14 CdtProblem
0.00/1.04 
↳15 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳16 BOUNDS(O(1), O(1))
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0.00/1.04 0.00/1.04
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0.00/1.04

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

2nd(cons(X, n__cons(Y, Z))) → activate(Y) 0.00/1.04
from(X) → cons(X, n__from(n__s(X))) 0.00/1.04
cons(X1, X2) → n__cons(X1, X2) 0.00/1.04
from(X) → n__from(X) 0.00/1.04
s(X) → n__s(X) 0.00/1.04
activate(n__cons(X1, X2)) → cons(activate(X1), X2) 0.00/1.04
activate(n__from(X)) → from(activate(X)) 0.00/1.04
activate(n__s(X)) → s(activate(X)) 0.00/1.04
activate(X) → X

Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

2nd(cons(z0, n__cons(z1, z2))) → activate(z1) 0.00/1.04
from(z0) → cons(z0, n__from(n__s(z0))) 0.00/1.04
from(z0) → n__from(z0) 0.00/1.04
cons(z0, z1) → n__cons(z0, z1) 0.00/1.04
s(z0) → n__s(z0) 0.00/1.04
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 0.00/1.04
activate(n__from(z0)) → from(activate(z0)) 0.00/1.04
activate(n__s(z0)) → s(activate(z0)) 0.00/1.04
activate(z0) → z0
Tuples:

2ND(cons(z0, n__cons(z1, z2))) → c(ACTIVATE(z1)) 0.00/1.04
FROM(z0) → c1(CONS(z0, n__from(n__s(z0)))) 0.00/1.04
ACTIVATE(n__cons(z0, z1)) → c5(CONS(activate(z0), z1), ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
S tuples:

2ND(cons(z0, n__cons(z1, z2))) → c(ACTIVATE(z1)) 0.00/1.04
FROM(z0) → c1(CONS(z0, n__from(n__s(z0)))) 0.00/1.04
ACTIVATE(n__cons(z0, z1)) → c5(CONS(activate(z0), z1), ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

2nd, from, cons, s, activate

Defined Pair Symbols:

2ND, FROM, ACTIVATE

Compound Symbols:

c, c1, c5, c6, c7

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(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

2ND(cons(z0, n__cons(z1, z2))) → c(ACTIVATE(z1))
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

2nd(cons(z0, n__cons(z1, z2))) → activate(z1) 0.00/1.04
from(z0) → cons(z0, n__from(n__s(z0))) 0.00/1.04
from(z0) → n__from(z0) 0.00/1.04
cons(z0, z1) → n__cons(z0, z1) 0.00/1.04
s(z0) → n__s(z0) 0.00/1.04
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 0.00/1.04
activate(n__from(z0)) → from(activate(z0)) 0.00/1.04
activate(n__s(z0)) → s(activate(z0)) 0.00/1.04
activate(z0) → z0
Tuples:

FROM(z0) → c1(CONS(z0, n__from(n__s(z0)))) 0.00/1.04
ACTIVATE(n__cons(z0, z1)) → c5(CONS(activate(z0), z1), ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
S tuples:

FROM(z0) → c1(CONS(z0, n__from(n__s(z0)))) 0.00/1.04
ACTIVATE(n__cons(z0, z1)) → c5(CONS(activate(z0), z1), ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

2nd, from, cons, s, activate

Defined Pair Symbols:

FROM, ACTIVATE

Compound Symbols:

c1, c5, c6, c7

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(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing tuple parts
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

2nd(cons(z0, n__cons(z1, z2))) → activate(z1) 0.00/1.04
from(z0) → cons(z0, n__from(n__s(z0))) 0.00/1.04
from(z0) → n__from(z0) 0.00/1.04
cons(z0, z1) → n__cons(z0, z1) 0.00/1.04
s(z0) → n__s(z0) 0.00/1.04
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 0.00/1.04
activate(n__from(z0)) → from(activate(z0)) 0.00/1.04
activate(n__s(z0)) → s(activate(z0)) 0.00/1.04
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.04
FROM(z0) → c1 0.00/1.04
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.04
FROM(z0) → c1 0.00/1.04
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.04
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

2nd, from, cons, s, activate

Defined Pair Symbols:

ACTIVATE, FROM

Compound Symbols:

c6, c1, c5, c7

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(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

FROM(z0) → c1
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

2nd(cons(z0, n__cons(z1, z2))) → activate(z1) 0.00/1.04
from(z0) → cons(z0, n__from(n__s(z0))) 0.00/1.04
from(z0) → n__from(z0) 0.00/1.04
cons(z0, z1) → n__cons(z0, z1) 0.00/1.04
s(z0) → n__s(z0) 0.00/1.04
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 0.00/1.04
activate(n__from(z0)) → from(activate(z0)) 0.00/1.04
activate(n__s(z0)) → s(activate(z0)) 0.00/1.04
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
FROM(z0) → c1 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
FROM(z0) → c1 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

2nd, from, cons, s, activate

Defined Pair Symbols:

ACTIVATE, FROM

Compound Symbols:

c6, c1, c5, c7

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(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0))
We considered the (Usable) Rules:

activate(n__cons(z0, z1)) → cons(activate(z0), z1) 0.00/1.05
activate(n__from(z0)) → from(activate(z0)) 0.00/1.05
activate(n__s(z0)) → s(activate(z0)) 0.00/1.05
activate(z0) → z0 0.00/1.05
s(z0) → n__s(z0) 0.00/1.05
from(z0) → cons(z0, n__from(n__s(z0))) 0.00/1.05
from(z0) → n__from(z0) 0.00/1.05
cons(z0, z1) → n__cons(z0, z1)
And the Tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
FROM(z0) → c1 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/1.05

POL(ACTIVATE(x1)) = [4]x1    0.00/1.05
POL(FROM(x1)) = 0    0.00/1.05
POL(activate(x1)) = 0    0.00/1.05
POL(c1) = 0    0.00/1.05
POL(c5(x1)) = x1    0.00/1.05
POL(c6(x1, x2)) = x1 + x2    0.00/1.05
POL(c7(x1)) = x1    0.00/1.05
POL(cons(x1, x2)) = [3] + [3]x1    0.00/1.05
POL(from(x1)) = [3] + [3]x1    0.00/1.05
POL(n__cons(x1, x2)) = [4] + x1 + x2    0.00/1.05
POL(n__from(x1)) = [1] + x1    0.00/1.05
POL(n__s(x1)) = x1    0.00/1.05
POL(s(x1)) = [3] + [3]x1   
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(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

2nd(cons(z0, n__cons(z1, z2))) → activate(z1) 0.00/1.05
from(z0) → cons(z0, n__from(n__s(z0))) 0.00/1.05
from(z0) → n__from(z0) 0.00/1.05
cons(z0, z1) → n__cons(z0, z1) 0.00/1.05
s(z0) → n__s(z0) 0.00/1.05
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 0.00/1.05
activate(n__from(z0)) → from(activate(z0)) 0.00/1.05
activate(n__s(z0)) → s(activate(z0)) 0.00/1.05
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
FROM(z0) → c1 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:

FROM(z0) → c1 0.00/1.05
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
K tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0))
Defined Rule Symbols:

2nd, from, cons, s, activate

Defined Pair Symbols:

ACTIVATE, FROM

Compound Symbols:

c6, c1, c5, c7

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(11) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

The following tuples could be moved from S to K by knowledge propagation:

FROM(z0) → c1
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(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

2nd(cons(z0, n__cons(z1, z2))) → activate(z1) 0.00/1.05
from(z0) → cons(z0, n__from(n__s(z0))) 0.00/1.05
from(z0) → n__from(z0) 0.00/1.05
cons(z0, z1) → n__cons(z0, z1) 0.00/1.05
s(z0) → n__s(z0) 0.00/1.05
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 0.00/1.05
activate(n__from(z0)) → from(activate(z0)) 0.00/1.05
activate(n__s(z0)) → s(activate(z0)) 0.00/1.05
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
FROM(z0) → c1 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:

ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
K tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.05
FROM(z0) → c1
Defined Rule Symbols:

2nd, from, cons, s, activate

Defined Pair Symbols:

ACTIVATE, FROM

Compound Symbols:

c6, c1, c5, c7

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(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
We considered the (Usable) Rules:

activate(n__cons(z0, z1)) → cons(activate(z0), z1) 0.00/1.05
activate(n__from(z0)) → from(activate(z0)) 0.00/1.05
activate(n__s(z0)) → s(activate(z0)) 0.00/1.05
activate(z0) → z0 0.00/1.05
s(z0) → n__s(z0) 0.00/1.05
from(z0) → cons(z0, n__from(n__s(z0))) 0.00/1.05
from(z0) → n__from(z0) 0.00/1.05
cons(z0, z1) → n__cons(z0, z1)
And the Tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
FROM(z0) → c1 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/1.05

POL(ACTIVATE(x1)) = [4]x1    0.00/1.05
POL(FROM(x1)) = [5]    0.00/1.05
POL(activate(x1)) = 0    0.00/1.05
POL(c1) = 0    0.00/1.05
POL(c5(x1)) = x1    0.00/1.05
POL(c6(x1, x2)) = x1 + x2    0.00/1.05
POL(c7(x1)) = x1    0.00/1.05
POL(cons(x1, x2)) = [3] + [3]x1    0.00/1.05
POL(from(x1)) = [3] + [3]x1    0.00/1.05
POL(n__cons(x1, x2)) = [3] + x1 + x2    0.00/1.05
POL(n__from(x1)) = [2] + x1    0.00/1.05
POL(n__s(x1)) = [1] + x1    0.00/1.05
POL(s(x1)) = [3] + [3]x1   
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(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

2nd(cons(z0, n__cons(z1, z2))) → activate(z1) 0.00/1.05
from(z0) → cons(z0, n__from(n__s(z0))) 0.00/1.05
from(z0) → n__from(z0) 0.00/1.05
cons(z0, z1) → n__cons(z0, z1) 0.00/1.05
s(z0) → n__s(z0) 0.00/1.05
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 0.00/1.05
activate(n__from(z0)) → from(activate(z0)) 0.00/1.05
activate(n__s(z0)) → s(activate(z0)) 0.00/1.05
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
FROM(z0) → c1 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:none
K tuples:

ACTIVATE(n__from(z0)) → c6(FROM(activate(z0)), ACTIVATE(z0)) 0.00/1.05
ACTIVATE(n__cons(z0, z1)) → c5(ACTIVATE(z0)) 0.00/1.05
FROM(z0) → c1 0.00/1.05
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
Defined Rule Symbols:

2nd, from, cons, s, activate

Defined Pair Symbols:

ACTIVATE, FROM

Compound Symbols:

c6, c1, c5, c7

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(15) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty
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(16) BOUNDS(O(1), O(1))

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2.80/1.25 EOF