YES(O(1), O(n^1)) 0.00/0.84 YES(O(1), O(n^1)) 0.00/0.90 0.00/0.90 0.00/0.90 0.00/0.90 0.00/0.90 0.00/0.90 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.90 0.00/0.90 0.00/0.90
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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
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↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳4 CdtProblem
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↳5 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳6 BOUNDS(O(1), O(1))
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0.00/0.90 0.00/0.90
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(0) Obligation:

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

++(nil, y) → y 0.00/0.90
++(x, nil) → x 0.00/0.90
++(.(x, y), z) → .(x, ++(y, z)) 0.00/0.90
++(++(x, y), z) → ++(x, ++(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:

++(nil, z0) → z0 0.00/0.90
++(z0, nil) → z0 0.00/0.90
++(.(z0, z1), z2) → .(z0, ++(z1, z2)) 0.00/0.90
++(++(z0, z1), z2) → ++(z0, ++(z1, z2))
Tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2)) 0.00/0.90
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))
S tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2)) 0.00/0.90
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))
K tuples:none
Defined Rule Symbols:

++

Defined Pair Symbols:

++'

Compound Symbols:

c2, c3

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(3) 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.

++'(.(z0, z1), z2) → c2(++'(z1, z2)) 0.00/0.90
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))
We considered the (Usable) Rules:

++(z0, nil) → z0 0.00/0.90
++(.(z0, z1), z2) → .(z0, ++(z1, z2)) 0.00/0.90
++(++(z0, z1), z2) → ++(z0, ++(z1, z2)) 0.00/0.90
++(nil, z0) → z0
And the Tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2)) 0.00/0.90
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.90

POL(++(x1, x2)) = [4] + [4]x1 + x2    0.00/0.90
POL(++'(x1, x2)) = [4]x1    0.00/0.90
POL(.(x1, x2)) = [1] + x1 + x2    0.00/0.90
POL(c2(x1)) = x1    0.00/0.90
POL(c3(x1, x2)) = x1 + x2    0.00/0.90
POL(nil) = 0   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

++(nil, z0) → z0 0.00/0.90
++(z0, nil) → z0 0.00/0.90
++(.(z0, z1), z2) → .(z0, ++(z1, z2)) 0.00/0.90
++(++(z0, z1), z2) → ++(z0, ++(z1, z2))
Tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2)) 0.00/0.90
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))
S tuples:none
K tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2)) 0.00/0.90
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))
Defined Rule Symbols:

++

Defined Pair Symbols:

++'

Compound Symbols:

c2, c3

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

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

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0.00/0.96 EOF