YES(O(1), O(n^1)) 0.00/0.74 YES(O(1), O(n^1)) 0.00/0.75 0.00/0.75 0.00/0.75 0.00/0.75 0.00/0.75 0.00/0.75 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.75 0.00/0.75 0.00/0.75
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(0) Obligation:

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

anchored(Cons(x, xs), y) → anchored(xs, Cons(Cons(Nil, Nil), y)) 0.00/0.75
anchored(Nil, y) → y 0.00/0.75
goal(x, y) → anchored(x, y)

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:

anchored(Cons(z0, z1), z2) → anchored(z1, Cons(Cons(Nil, Nil), z2)) 0.00/0.75
anchored(Nil, z0) → z0 0.00/0.75
goal(z0, z1) → anchored(z0, z1)
Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2))) 0.00/0.75
GOAL(z0, z1) → c2(ANCHORED(z0, z1))
S tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2))) 0.00/0.75
GOAL(z0, z1) → c2(ANCHORED(z0, z1))
K tuples:none
Defined Rule Symbols:

anchored, goal

Defined Pair Symbols:

ANCHORED, GOAL

Compound Symbols:

c, c2

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(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c2(ANCHORED(z0, z1))
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

anchored(Cons(z0, z1), z2) → anchored(z1, Cons(Cons(Nil, Nil), z2)) 0.00/0.75
anchored(Nil, z0) → z0 0.00/0.75
goal(z0, z1) → anchored(z0, z1)
Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
S tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
K tuples:none
Defined Rule Symbols:

anchored, goal

Defined Pair Symbols:

ANCHORED

Compound Symbols:

c

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(5) 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.

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
We considered the (Usable) Rules:none
And the Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.75

POL(ANCHORED(x1, x2)) = [4]x1 + [3]x2    0.00/0.75
POL(Cons(x1, x2)) = [4] + x2    0.00/0.75
POL(Nil) = 0    0.00/0.75
POL(c(x1)) = x1   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

anchored(Cons(z0, z1), z2) → anchored(z1, Cons(Cons(Nil, Nil), z2)) 0.00/0.75
anchored(Nil, z0) → z0 0.00/0.75
goal(z0, z1) → anchored(z0, z1)
Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
S tuples:none
K tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
Defined Rule Symbols:

anchored, goal

Defined Pair Symbols:

ANCHORED

Compound Symbols:

c

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

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

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0.00/0.77 EOF