YES(O(1), O(n^1)) 0.00/0.78 YES(O(1), O(n^1)) 0.00/0.80 0.00/0.80 0.00/0.80 0.00/0.80 0.00/0.80 0.00/0.80 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.80 0.00/0.80 0.00/0.80
0.00/0.80
0.00/0.80
0.00/0.80
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
0.00/0.80 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/0.80 
↳2 CdtProblem
0.00/0.80 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
0.00/0.80 
↳4 CdtProblem
0.00/0.80 
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
0.00/0.80 
↳6 CdtProblem
0.00/0.80 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.80 
↳8 CdtProblem
0.00/0.80 
↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/0.80 
↳10 BOUNDS(O(1), O(1))
0.00/0.80
0.00/0.80 0.00/0.80
0.00/0.80
0.00/0.80

(0) Obligation:

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

from(X) → cons(X, n__from(s(X))) 0.00/0.80
sel(0, cons(X, Y)) → X 0.00/0.80
sel(s(X), cons(Y, Z)) → sel(X, activate(Z)) 0.00/0.80
from(X) → n__from(X) 0.00/0.80
activate(n__from(X)) → from(X) 0.00/0.80
activate(X) → X

Rewrite Strategy: INNERMOST
0.00/0.80
0.00/0.80

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

Converted CpxTRS to CDT
0.00/0.80
0.00/0.80

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 0.00/0.80
from(z0) → n__from(z0) 0.00/0.80
sel(0, cons(z0, z1)) → z0 0.00/0.80
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2)) 0.00/0.80
activate(n__from(z0)) → from(z0) 0.00/0.80
activate(z0) → z0
Tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4(FROM(z0))
S tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4(FROM(z0))
K tuples:none
Defined Rule Symbols:

from, sel, activate

Defined Pair Symbols:

SEL, ACTIVATE

Compound Symbols:

c3, c4

0.00/0.80
0.00/0.80

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

Removed 1 trailing tuple parts
0.00/0.80
0.00/0.80

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 0.00/0.80
from(z0) → n__from(z0) 0.00/0.80
sel(0, cons(z0, z1)) → z0 0.00/0.80
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2)) 0.00/0.80
activate(n__from(z0)) → from(z0) 0.00/0.80
activate(z0) → z0
Tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4
S tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4
K tuples:none
Defined Rule Symbols:

from, sel, activate

Defined Pair Symbols:

SEL, ACTIVATE

Compound Symbols:

c3, c4

0.00/0.80
0.00/0.80

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

Removed 1 trailing nodes:

ACTIVATE(n__from(z0)) → c4
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0.00/0.80

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 0.00/0.80
from(z0) → n__from(z0) 0.00/0.80
sel(0, cons(z0, z1)) → z0 0.00/0.80
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2)) 0.00/0.80
activate(n__from(z0)) → from(z0) 0.00/0.80
activate(z0) → z0
Tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4
S tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4
K tuples:none
Defined Rule Symbols:

from, sel, activate

Defined Pair Symbols:

SEL, ACTIVATE

Compound Symbols:

c3, c4

0.00/0.80
0.00/0.80

(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)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4
We considered the (Usable) Rules:

activate(n__from(z0)) → from(z0) 0.00/0.80
activate(z0) → z0 0.00/0.80
from(z0) → cons(z0, n__from(s(z0))) 0.00/0.80
from(z0) → n__from(z0)
And the Tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.80

POL(ACTIVATE(x1)) = [1]    0.00/0.80
POL(SEL(x1, x2)) = [2]x1    0.00/0.80
POL(activate(x1)) = 0    0.00/0.80
POL(c3(x1, x2)) = x1 + x2    0.00/0.80
POL(c4) = 0    0.00/0.80
POL(cons(x1, x2)) = x1    0.00/0.80
POL(from(x1)) = [3] + [3]x1    0.00/0.80
POL(n__from(x1)) = x1    0.00/0.80
POL(s(x1)) = [1] + x1   
0.00/0.80
0.00/0.80

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 0.00/0.80
from(z0) → n__from(z0) 0.00/0.80
sel(0, cons(z0, z1)) → z0 0.00/0.80
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2)) 0.00/0.80
activate(n__from(z0)) → from(z0) 0.00/0.80
activate(z0) → z0
Tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4
S tuples:none
K tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.80
ACTIVATE(n__from(z0)) → c4
Defined Rule Symbols:

from, sel, activate

Defined Pair Symbols:

SEL, ACTIVATE

Compound Symbols:

c3, c4

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0.00/0.80

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

The set S is empty
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0.00/0.80

(10) BOUNDS(O(1), O(1))

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0.00/0.80
0.00/0.84 EOF