YES(O(1), O(n^2)) 2.80/1.11 YES(O(1), O(n^2)) 2.80/1.14 2.80/1.14 2.80/1.14 2.80/1.14 2.80/1.14 2.80/1.14 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 2.80/1.14 2.80/1.14 2.80/1.14
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
2.80/1.14 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
2.80/1.14 
↳2 CdtProblem
2.80/1.14 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
2.80/1.14 
↳4 CdtProblem
2.80/1.14 
↳5 CdtLeafRemovalProof (ComplexityIfPolyImplication)
2.80/1.14 
↳6 CdtProblem
2.80/1.14 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
2.80/1.14 
↳8 CdtProblem
2.80/1.14 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
2.80/1.14 
↳10 CdtProblem
2.80/1.14 
↳11 CdtKnowledgeProof (⇔)
2.80/1.14 
↳12 BOUNDS(O(1), O(1))
2.80/1.14
2.80/1.14 2.80/1.14
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2.80/1.14

(0) Obligation:

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

from(X) → cons(X, n__from(s(X))) 2.80/1.14
head(cons(X, XS)) → X 2.80/1.14
2nd(cons(X, XS)) → head(activate(XS)) 2.80/1.14
take(0, XS) → nil 2.80/1.14
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS))) 2.80/1.14
sel(0, cons(X, XS)) → X 2.80/1.14
sel(s(N), cons(X, XS)) → sel(N, activate(XS)) 2.80/1.14
from(X) → n__from(X) 2.80/1.14
take(X1, X2) → n__take(X1, X2) 2.80/1.14
activate(n__from(X)) → from(X) 2.80/1.14
activate(n__take(X1, X2)) → take(X1, X2) 2.80/1.14
activate(X) → X

Rewrite Strategy: INNERMOST
2.80/1.14
2.80/1.14

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
2.80/1.14
2.80/1.14

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 2.80/1.14
from(z0) → n__from(z0) 2.80/1.14
head(cons(z0, z1)) → z0 2.80/1.14
2nd(cons(z0, z1)) → head(activate(z1)) 2.80/1.14
take(0, z0) → nil 2.80/1.14
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2))) 2.80/1.14
take(z0, z1) → n__take(z0, z1) 2.80/1.14
sel(0, cons(z0, z1)) → z0 2.80/1.14
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2)) 2.80/1.14
activate(n__from(z0)) → from(z0) 2.80/1.14
activate(n__take(z0, z1)) → take(z0, z1) 2.80/1.14
activate(z0) → z0
Tuples:

2ND(cons(z0, z1)) → c3(HEAD(activate(z1)), ACTIVATE(z1)) 2.80/1.14
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9(FROM(z0)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
S tuples:

2ND(cons(z0, z1)) → c3(HEAD(activate(z1)), ACTIVATE(z1)) 2.80/1.14
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9(FROM(z0)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
K tuples:none
Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

2ND, TAKE, SEL, ACTIVATE

Compound Symbols:

c3, c5, c8, c9, c10

2.80/1.14
2.80/1.14

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

Removed 2 trailing tuple parts
2.80/1.14
2.80/1.14

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 2.80/1.14
from(z0) → n__from(z0) 2.80/1.14
head(cons(z0, z1)) → z0 2.80/1.14
2nd(cons(z0, z1)) → head(activate(z1)) 2.80/1.14
take(0, z0) → nil 2.80/1.14
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2))) 2.80/1.14
take(z0, z1) → n__take(z0, z1) 2.80/1.14
sel(0, cons(z0, z1)) → z0 2.80/1.14
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2)) 2.80/1.14
activate(n__from(z0)) → from(z0) 2.80/1.14
activate(n__take(z0, z1)) → take(z0, z1) 2.80/1.14
activate(z0) → z0
Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1)) 2.80/1.14
2ND(cons(z0, z1)) → c3(ACTIVATE(z1)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
S tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1)) 2.80/1.14
2ND(cons(z0, z1)) → c3(ACTIVATE(z1)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
K tuples:none
Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

TAKE, SEL, ACTIVATE, 2ND

Compound Symbols:

c5, c8, c10, c3, c9

2.80/1.14
2.80/1.14

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

2ND(cons(z0, z1)) → c3(ACTIVATE(z1))
Removed 1 trailing nodes:

ACTIVATE(n__from(z0)) → c9
2.80/1.14
2.80/1.14

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 2.80/1.14
from(z0) → n__from(z0) 2.80/1.14
head(cons(z0, z1)) → z0 2.80/1.14
2nd(cons(z0, z1)) → head(activate(z1)) 2.80/1.14
take(0, z0) → nil 2.80/1.14
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2))) 2.80/1.14
take(z0, z1) → n__take(z0, z1) 2.80/1.14
sel(0, cons(z0, z1)) → z0 2.80/1.14
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2)) 2.80/1.14
activate(n__from(z0)) → from(z0) 2.80/1.14
activate(n__take(z0, z1)) → take(z0, z1) 2.80/1.14
activate(z0) → z0
Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
S tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
K tuples:none
Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

TAKE, SEL, ACTIVATE

Compound Symbols:

c5, c8, c10, c9

2.80/1.14
2.80/1.14

(7) 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.

SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
We considered the (Usable) Rules:

activate(n__from(z0)) → from(z0) 2.80/1.14
activate(n__take(z0, z1)) → take(z0, z1) 2.80/1.14
activate(z0) → z0 2.80/1.14
take(0, z0) → nil 2.80/1.14
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2))) 2.80/1.14
take(z0, z1) → n__take(z0, z1) 2.80/1.14
from(z0) → cons(z0, n__from(s(z0))) 2.80/1.14
from(z0) → n__from(z0)
And the Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
The order we found is given by the following interpretation:
Polynomial interpretation : 2.80/1.14

POL(0) = [3]    2.80/1.14
POL(ACTIVATE(x1)) = [4]    2.80/1.14
POL(SEL(x1, x2)) = [4]x1    2.80/1.14
POL(TAKE(x1, x2)) = [4]    2.80/1.14
POL(activate(x1)) = [4]x1    2.80/1.14
POL(c10(x1)) = x1    2.80/1.14
POL(c5(x1)) = x1    2.80/1.14
POL(c8(x1, x2)) = x1 + x2    2.80/1.14
POL(c9) = 0    2.80/1.14
POL(cons(x1, x2)) = [4]    2.80/1.14
POL(from(x1)) = [4] + [2]x1    2.80/1.14
POL(n__from(x1)) = [2] + x1    2.80/1.14
POL(n__take(x1, x2)) = [2] + x2    2.80/1.14
POL(nil) = [1]    2.80/1.14
POL(s(x1)) = [2] + x1    2.80/1.14
POL(take(x1, x2)) = [2] + [2]x2   
2.80/1.14
2.80/1.14

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 2.80/1.14
from(z0) → n__from(z0) 2.80/1.14
head(cons(z0, z1)) → z0 2.80/1.14
2nd(cons(z0, z1)) → head(activate(z1)) 2.80/1.14
take(0, z0) → nil 2.80/1.14
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2))) 2.80/1.14
take(z0, z1) → n__take(z0, z1) 2.80/1.14
sel(0, cons(z0, z1)) → z0 2.80/1.14
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2)) 2.80/1.14
activate(n__from(z0)) → from(z0) 2.80/1.14
activate(n__take(z0, z1)) → take(z0, z1) 2.80/1.14
activate(z0) → z0
Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
S tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
K tuples:

SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

TAKE, SEL, ACTIVATE

Compound Symbols:

c5, c8, c10, c9

2.80/1.14
2.80/1.14

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
We considered the (Usable) Rules:

activate(n__from(z0)) → from(z0) 2.80/1.14
activate(n__take(z0, z1)) → take(z0, z1) 2.80/1.14
activate(z0) → z0 2.80/1.14
take(0, z0) → nil 2.80/1.14
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2))) 2.80/1.14
take(z0, z1) → n__take(z0, z1) 2.80/1.14
from(z0) → cons(z0, n__from(s(z0))) 2.80/1.14
from(z0) → n__from(z0)
And the Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
The order we found is given by the following interpretation:
Polynomial interpretation : 2.80/1.14

POL(0) = [3]    2.80/1.14
POL(ACTIVATE(x1)) = [2]x1    2.80/1.14
POL(SEL(x1, x2)) = x1·x2    2.80/1.14
POL(TAKE(x1, x2)) = [2]x2    2.80/1.14
POL(activate(x1)) = [1] + x1    2.80/1.14
POL(c10(x1)) = x1    2.80/1.14
POL(c5(x1)) = x1    2.80/1.14
POL(c8(x1, x2)) = x1 + x2    2.80/1.14
POL(c9) = 0    2.80/1.14
POL(cons(x1, x2)) = [1] + x2    2.80/1.14
POL(from(x1)) = [2]    2.80/1.14
POL(n__from(x1)) = [1]    2.80/1.14
POL(n__take(x1, x2)) = x2    2.80/1.14
POL(nil) = 0    2.80/1.14
POL(s(x1)) = [2] + x1    2.80/1.14
POL(take(x1, x2)) = [1] + x2   
2.80/1.14
2.80/1.14

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 2.80/1.14
from(z0) → n__from(z0) 2.80/1.14
head(cons(z0, z1)) → z0 2.80/1.14
2nd(cons(z0, z1)) → head(activate(z1)) 2.80/1.14
take(0, z0) → nil 2.80/1.14
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2))) 2.80/1.14
take(z0, z1) → n__take(z0, z1) 2.80/1.14
sel(0, cons(z0, z1)) → z0 2.80/1.14
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2)) 2.80/1.14
activate(n__from(z0)) → from(z0) 2.80/1.14
activate(n__take(z0, z1)) → take(z0, z1) 2.80/1.14
activate(z0) → z0
Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2)) 2.80/1.14
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9
S tuples:

ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
K tuples:

SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2)) 2.80/1.14
ACTIVATE(n__from(z0)) → c9 2.80/1.14
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

TAKE, SEL, ACTIVATE

Compound Symbols:

c5, c8, c10, c9

2.80/1.14
2.80/1.14

(11) CdtKnowledgeProof (EQUIVALENT transformation)

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

ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1)) 2.80/1.14
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
Now S is empty
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2.80/1.14

(12) BOUNDS(O(1), O(1))

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2.80/1.14
3.03/1.23 EOF