YES(O(1), O(n^2)) 0.00/0.85 YES(O(1), O(n^2)) 0.00/0.87 0.00/0.87 0.00/0.87 0.00/0.87 0.00/0.87 0.00/0.87 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.87 0.00/0.87 0.00/0.87
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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
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↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
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↳2 CdtProblem
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↳3 CdtUnreachableProof (⇔)
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↳4 CdtProblem
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↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
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↳6 CdtProblem
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↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
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↳8 CdtProblem
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↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳10 CdtProblem
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↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳12 CdtProblem
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↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳14 BOUNDS(O(1), O(1))
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0.00/0.87 0.00/0.87
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(0) Obligation:

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

f(cons(nil, y)) → y 0.00/0.87
f(cons(f(cons(nil, y)), z)) → copy(n, y, z) 0.00/0.87
copy(0, y, z) → f(z) 0.00/0.87
copy(s(x), y, z) → copy(x, y, cons(f(y), z))

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:

f(cons(nil, z0)) → z0 0.00/0.87
f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1) 0.00/0.87
copy(0, z0, z1) → f(z1) 0.00/0.87
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))
Tuples:

F(cons(f(cons(nil, z0)), z1)) → c1(COPY(n, z0, z1)) 0.00/0.87
COPY(0, z0, z1) → c2(F(z1)) 0.00/0.87
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)), F(z1))
S tuples:

F(cons(f(cons(nil, z0)), z1)) → c1(COPY(n, z0, z1)) 0.00/0.87
COPY(0, z0, z1) → c2(F(z1)) 0.00/0.87
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)), F(z1))
K tuples:none
Defined Rule Symbols:

f, copy

Defined Pair Symbols:

F, COPY

Compound Symbols:

c1, c2, c3

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(3) CdtUnreachableProof (EQUIVALENT transformation)

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

F(cons(f(cons(nil, z0)), z1)) → c1(COPY(n, z0, z1))
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0 0.00/0.88
f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1) 0.00/0.88
copy(0, z0, z1) → f(z1) 0.00/0.88
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))
Tuples:

COPY(0, z0, z1) → c2(F(z1)) 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)), F(z1))
S tuples:

COPY(0, z0, z1) → c2(F(z1)) 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)), F(z1))
K tuples:none
Defined Rule Symbols:

f, copy

Defined Pair Symbols:

COPY

Compound Symbols:

c2, c3

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

Removed 2 trailing tuple parts
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0 0.00/0.88
f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1) 0.00/0.88
copy(0, z0, z1) → f(z1) 0.00/0.88
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))
Tuples:

COPY(0, z0, z1) → c2 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
S tuples:

COPY(0, z0, z1) → c2 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
K tuples:none
Defined Rule Symbols:

f, copy

Defined Pair Symbols:

COPY

Compound Symbols:

c2, c3

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

Removed 1 trailing nodes:

COPY(0, z0, z1) → c2
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0 0.00/0.88
f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1) 0.00/0.88
copy(0, z0, z1) → f(z1) 0.00/0.88
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))
Tuples:

COPY(0, z0, z1) → c2 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
S tuples:

COPY(0, z0, z1) → c2 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
K tuples:none
Defined Rule Symbols:

f, copy

Defined Pair Symbols:

COPY

Compound Symbols:

c2, c3

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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.

COPY(0, z0, z1) → c2
We considered the (Usable) Rules:

f(cons(nil, z0)) → z0
And the Tuples:

COPY(0, z0, z1) → c2 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.88

POL(0) = [3]    0.00/0.88
POL(COPY(x1, x2, x3)) = [5]    0.00/0.88
POL(c2) = 0    0.00/0.88
POL(c3(x1)) = x1    0.00/0.88
POL(cons(x1, x2)) = [1] + x2    0.00/0.88
POL(f(x1)) = [1]    0.00/0.88
POL(nil) = 0    0.00/0.88
POL(s(x1)) = [1] + x1   
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(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0 0.00/0.88
f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1) 0.00/0.88
copy(0, z0, z1) → f(z1) 0.00/0.88
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))
Tuples:

COPY(0, z0, z1) → c2 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
S tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
K tuples:

COPY(0, z0, z1) → c2
Defined Rule Symbols:

f, copy

Defined Pair Symbols:

COPY

Compound Symbols:

c2, c3

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(11) 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.

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
We considered the (Usable) Rules:

f(cons(nil, z0)) → z0
And the Tuples:

COPY(0, z0, z1) → c2 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.88

POL(0) = 0    0.00/0.88
POL(COPY(x1, x2, x3)) = [2]x3 + x12    0.00/0.88
POL(c2) = 0    0.00/0.88
POL(c3(x1)) = x1    0.00/0.88
POL(cons(x1, x2)) = 0    0.00/0.88
POL(f(x1)) = [3]x1    0.00/0.88
POL(nil) = [3]    0.00/0.88
POL(s(x1)) = [2] + x1   
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(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0 0.00/0.88
f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1) 0.00/0.88
copy(0, z0, z1) → f(z1) 0.00/0.88
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))
Tuples:

COPY(0, z0, z1) → c2 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
S tuples:none
K tuples:

COPY(0, z0, z1) → c2 0.00/0.88
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
Defined Rule Symbols:

f, copy

Defined Pair Symbols:

COPY

Compound Symbols:

c2, c3

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

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

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0.00/0.93 EOF