YES(O(1), O(n^1)) 0.00/0.88 YES(O(1), O(n^1)) 0.00/0.91 0.00/0.91 0.00/0.91 0.00/0.91 0.00/0.91 0.00/0.91 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.91 0.00/0.91 0.00/0.91
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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 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
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↳4 CdtProblem
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↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
0.00/0.91 
↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳8 CdtProblem
0.00/0.91 
↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/0.91 
↳10 BOUNDS(O(1), O(1))
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0.00/0.91 0.00/0.91
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0.00/0.91

(0) Obligation:

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

:(:(x, y), z) → :(x, :(y, z)) 0.00/0.91
:(+(x, y), z) → +(:(x, z), :(y, z)) 0.00/0.91
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
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0.00/0.91

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

:(:(z0, z1), z2) → :(z0, :(z1, z2)) 0.00/0.91
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2)) 0.00/0.91
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2(:'(g(z0, z2), +(z1, a)))
S tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2(:'(g(z0, z2), +(z1, a)))
K tuples:none
Defined Rule Symbols:

:

Defined Pair Symbols:

:'

Compound Symbols:

c, c1, c2

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

Removed 1 trailing tuple parts
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0.00/0.91

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

:(:(z0, z1), z2) → :(z0, :(z1, z2)) 0.00/0.91
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2)) 0.00/0.91
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2
S tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2
K tuples:none
Defined Rule Symbols:

:

Defined Pair Symbols:

:'

Compound Symbols:

c, c1, c2

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0.00/0.91

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

Removed 1 trailing nodes:

:'(z0, +(z1, f(z2))) → c2
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0.00/0.91

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

:(:(z0, z1), z2) → :(z0, :(z1, z2)) 0.00/0.91
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2)) 0.00/0.91
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2
S tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2
K tuples:none
Defined Rule Symbols:

:

Defined Pair Symbols:

:'

Compound Symbols:

c, c1, c2

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0.00/0.91

(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.

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2
We considered the (Usable) Rules:

:(:(z0, z1), z2) → :(z0, :(z1, z2)) 0.00/0.91
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2)) 0.00/0.91
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
And the Tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.91

POL(+(x1, x2)) = [4] + x1 + x2    0.00/0.91
POL(:(x1, x2)) = [4] + [4]x1 + x2    0.00/0.91
POL(:'(x1, x2)) = [4] + [4]x1    0.00/0.91
POL(a) = [3]    0.00/0.91
POL(c(x1, x2)) = x1 + x2    0.00/0.91
POL(c1(x1, x2)) = x1 + x2    0.00/0.91
POL(c2) = 0    0.00/0.91
POL(f(x1)) = [3] + x1    0.00/0.91
POL(g(x1, x2)) = [3] + x1 + x2   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

:(:(z0, z1), z2) → :(z0, :(z1, z2)) 0.00/0.91
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2)) 0.00/0.91
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2
S tuples:none
K tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2)) 0.00/0.91
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2)) 0.00/0.91
:'(z0, +(z1, f(z2))) → c2
Defined Rule Symbols:

:

Defined Pair Symbols:

:'

Compound Symbols:

c, c1, c2

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

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

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0.00/0.95 EOF