YES(O(1), O(n^1)) 0.00/0.76 YES(O(1), O(n^1)) 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.77 0.00/0.77 0.00/0.77
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(0) Obligation:

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

a__f(X) → g(h(f(X))) 0.00/0.77
mark(f(X)) → a__f(mark(X)) 0.00/0.77
mark(g(X)) → g(X) 0.00/0.77
mark(h(X)) → h(mark(X)) 0.00/0.77
a__f(X) → f(X)

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

Converted CpxTRS to CDT
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0.00/0.77

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0) → g(h(f(z0))) 0.00/0.77
a__f(z0) → f(z0) 0.00/0.77
mark(f(z0)) → a__f(mark(z0)) 0.00/0.77
mark(g(z0)) → g(z0) 0.00/0.77
mark(h(z0)) → h(mark(z0))
Tuples:

MARK(f(z0)) → c2(A__F(mark(z0)), MARK(z0)) 0.00/0.77
MARK(h(z0)) → c4(MARK(z0))
S tuples:

MARK(f(z0)) → c2(A__F(mark(z0)), MARK(z0)) 0.00/0.77
MARK(h(z0)) → c4(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c2, c4

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

Removed 1 trailing tuple parts
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0) → g(h(f(z0))) 0.00/0.77
a__f(z0) → f(z0) 0.00/0.77
mark(f(z0)) → a__f(mark(z0)) 0.00/0.77
mark(g(z0)) → g(z0) 0.00/0.77
mark(h(z0)) → h(mark(z0))
Tuples:

MARK(h(z0)) → c4(MARK(z0)) 0.00/0.77
MARK(f(z0)) → c2(MARK(z0))
S tuples:

MARK(h(z0)) → c4(MARK(z0)) 0.00/0.77
MARK(f(z0)) → c2(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c4, c2

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0.00/0.77

(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.

MARK(h(z0)) → c4(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(h(z0)) → c4(MARK(z0)) 0.00/0.77
MARK(f(z0)) → c2(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.77

POL(MARK(x1)) = [2]x1    0.00/0.77
POL(c2(x1)) = x1    0.00/0.77
POL(c4(x1)) = x1    0.00/0.77
POL(f(x1)) = x1    0.00/0.77
POL(h(x1)) = [1] + x1   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0) → g(h(f(z0))) 0.00/0.77
a__f(z0) → f(z0) 0.00/0.77
mark(f(z0)) → a__f(mark(z0)) 0.00/0.77
mark(g(z0)) → g(z0) 0.00/0.77
mark(h(z0)) → h(mark(z0))
Tuples:

MARK(h(z0)) → c4(MARK(z0)) 0.00/0.77
MARK(f(z0)) → c2(MARK(z0))
S tuples:

MARK(f(z0)) → c2(MARK(z0))
K tuples:

MARK(h(z0)) → c4(MARK(z0))
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c4, c2

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0.00/0.77

(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.

MARK(f(z0)) → c2(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(h(z0)) → c4(MARK(z0)) 0.00/0.77
MARK(f(z0)) → c2(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.77

POL(MARK(x1)) = [3]x1    0.00/0.77
POL(c2(x1)) = x1    0.00/0.77
POL(c4(x1)) = x1    0.00/0.77
POL(f(x1)) = [1] + x1    0.00/0.77
POL(h(x1)) = x1   
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0.00/0.77

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0) → g(h(f(z0))) 0.00/0.77
a__f(z0) → f(z0) 0.00/0.77
mark(f(z0)) → a__f(mark(z0)) 0.00/0.77
mark(g(z0)) → g(z0) 0.00/0.77
mark(h(z0)) → h(mark(z0))
Tuples:

MARK(h(z0)) → c4(MARK(z0)) 0.00/0.77
MARK(f(z0)) → c2(MARK(z0))
S tuples:none
K tuples:

MARK(h(z0)) → c4(MARK(z0)) 0.00/0.77
MARK(f(z0)) → c2(MARK(z0))
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c4, 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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0.00/0.78 EOF