YES(O(1), O(n^1)) 0.00/0.77 YES(O(1), O(n^1)) 0.00/0.78 0.00/0.78 0.00/0.78 0.00/0.78 0.00/0.78 0.00/0.78 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.78 0.00/0.78 0.00/0.78
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(0) Obligation:

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

a__f(f(a)) → c(f(g(f(a)))) 0.00/0.78
mark(f(X)) → a__f(mark(X)) 0.00/0.78
mark(a) → a 0.00/0.78
mark(c(X)) → c(X) 0.00/0.78
mark(g(X)) → g(mark(X)) 0.00/0.78
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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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(a)) → c(f(g(f(a)))) 0.00/0.78
a__f(z0) → f(z0) 0.00/0.78
mark(f(z0)) → a__f(mark(z0)) 0.00/0.78
mark(a) → a 0.00/0.78
mark(c(z0)) → c(z0) 0.00/0.78
mark(g(z0)) → g(mark(z0))
Tuples:

MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0)) 0.00/0.78
MARK(g(z0)) → c6(MARK(z0))
S tuples:

MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0)) 0.00/0.78
MARK(g(z0)) → c6(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c3, c6

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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(f(a)) → c(f(g(f(a)))) 0.00/0.78
a__f(z0) → f(z0) 0.00/0.78
mark(f(z0)) → a__f(mark(z0)) 0.00/0.78
mark(a) → a 0.00/0.78
mark(c(z0)) → c(z0) 0.00/0.78
mark(g(z0)) → g(mark(z0))
Tuples:

MARK(g(z0)) → c6(MARK(z0)) 0.00/0.78
MARK(f(z0)) → c3(MARK(z0))
S tuples:

MARK(g(z0)) → c6(MARK(z0)) 0.00/0.78
MARK(f(z0)) → c3(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c6, c3

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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.

MARK(g(z0)) → c6(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(g(z0)) → c6(MARK(z0)) 0.00/0.78
MARK(f(z0)) → c3(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.78

POL(MARK(x1)) = [2]x1    0.00/0.78
POL(c3(x1)) = x1    0.00/0.78
POL(c6(x1)) = x1    0.00/0.78
POL(f(x1)) = x1    0.00/0.78
POL(g(x1)) = [1] + x1   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(a)) → c(f(g(f(a)))) 0.00/0.78
a__f(z0) → f(z0) 0.00/0.78
mark(f(z0)) → a__f(mark(z0)) 0.00/0.78
mark(a) → a 0.00/0.78
mark(c(z0)) → c(z0) 0.00/0.78
mark(g(z0)) → g(mark(z0))
Tuples:

MARK(g(z0)) → c6(MARK(z0)) 0.00/0.78
MARK(f(z0)) → c3(MARK(z0))
S tuples:

MARK(f(z0)) → c3(MARK(z0))
K tuples:

MARK(g(z0)) → c6(MARK(z0))
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c6, c3

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0.00/0.78

(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)) → c3(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(g(z0)) → c6(MARK(z0)) 0.00/0.78
MARK(f(z0)) → c3(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.78

POL(MARK(x1)) = [3]x1    0.00/0.78
POL(c3(x1)) = x1    0.00/0.78
POL(c6(x1)) = x1    0.00/0.78
POL(f(x1)) = [1] + x1    0.00/0.78
POL(g(x1)) = x1   
0.00/0.78
0.00/0.78

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(a)) → c(f(g(f(a)))) 0.00/0.78
a__f(z0) → f(z0) 0.00/0.78
mark(f(z0)) → a__f(mark(z0)) 0.00/0.78
mark(a) → a 0.00/0.78
mark(c(z0)) → c(z0) 0.00/0.78
mark(g(z0)) → g(mark(z0))
Tuples:

MARK(g(z0)) → c6(MARK(z0)) 0.00/0.78
MARK(f(z0)) → c3(MARK(z0))
S tuples:none
K tuples:

MARK(g(z0)) → c6(MARK(z0)) 0.00/0.78
MARK(f(z0)) → c3(MARK(z0))
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c6, c3

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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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0.00/0.83 EOF