YES(O(1), O(n^1)) 0.00/0.96 YES(O(1), O(n^1)) 0.00/0.99 0.00/0.99 0.00/0.99 0.00/0.99 0.00/0.99 0.00/0.99 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.99 0.00/0.99 0.00/0.99
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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))
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↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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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 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳12 BOUNDS(O(1), O(1))
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(0) Obligation:

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

a__f(b, X, c) → a__f(X, a__c, X) 0.00/0.99
a__c → b 0.00/0.99
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3) 0.00/0.99
mark(c) → a__c 0.00/0.99
mark(b) → b 0.00/0.99
a__f(X1, X2, X3) → f(X1, X2, X3) 0.00/0.99
a__c → c

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(b, z0, c) → a__f(z0, a__c, z0) 0.00/0.99
a__f(z0, z1, z2) → f(z0, z1, z2) 0.00/0.99
a__c → b 0.00/0.99
a__c → c 0.00/0.99
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2) 0.00/0.99
mark(c) → a__c 0.00/0.99
mark(b) → b
Tuples:

A__F(b, z0, c) → c1(A__F(z0, a__c, z0), A__C) 0.00/0.99
MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
MARK(c) → c6(A__C)
S tuples:

A__F(b, z0, c) → c1(A__F(z0, a__c, z0), A__C) 0.00/0.99
MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
MARK(c) → c6(A__C)
K tuples:none
Defined Rule Symbols:

a__f, a__c, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c1, c5, c6

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

Removed 3 trailing tuple parts
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(b, z0, c) → a__f(z0, a__c, z0) 0.00/0.99
a__f(z0, z1, z2) → f(z0, z1, z2) 0.00/0.99
a__c → b 0.00/0.99
a__c → c 0.00/0.99
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2) 0.00/0.99
mark(c) → a__c 0.00/0.99
mark(b) → b
Tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1 0.00/0.99
MARK(c) → c6
S tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1 0.00/0.99
MARK(c) → c6
K tuples:none
Defined Rule Symbols:

a__f, a__c, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c5, c1, c6

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

Removed 2 trailing nodes:

MARK(c) → c6 0.00/0.99
A__F(b, z0, c) → c1
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(b, z0, c) → a__f(z0, a__c, z0) 0.00/0.99
a__f(z0, z1, z2) → f(z0, z1, z2) 0.00/0.99
a__c → b 0.00/0.99
a__c → c 0.00/0.99
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2) 0.00/0.99
mark(c) → a__c 0.00/0.99
mark(b) → b
Tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1 0.00/0.99
MARK(c) → c6
S tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1 0.00/0.99
MARK(c) → c6
K tuples:none
Defined Rule Symbols:

a__f, a__c, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c5, c1, c6

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(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(c) → c6
We considered the (Usable) Rules:

mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2) 0.00/0.99
mark(c) → a__c 0.00/0.99
mark(b) → b 0.00/0.99
a__c → b 0.00/0.99
a__c → c 0.00/0.99
a__f(b, z0, c) → a__f(z0, a__c, z0) 0.00/0.99
a__f(z0, z1, z2) → f(z0, z1, z2)
And the Tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1 0.00/0.99
MARK(c) → c6
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.99

POL(A__F(x1, x2, x3)) = 0    0.00/0.99
POL(MARK(x1)) = [3]    0.00/0.99
POL(a__c) = 0    0.00/0.99
POL(a__f(x1, x2, x3)) = [3] + [3]x2    0.00/0.99
POL(b) = [3]    0.00/0.99
POL(c) = [3]    0.00/0.99
POL(c1) = 0    0.00/0.99
POL(c5(x1, x2)) = x1 + x2    0.00/0.99
POL(c6) = 0    0.00/0.99
POL(f(x1, x2, x3)) = x1 + x3    0.00/0.99
POL(mark(x1)) = 0   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(b, z0, c) → a__f(z0, a__c, z0) 0.00/0.99
a__f(z0, z1, z2) → f(z0, z1, z2) 0.00/0.99
a__c → b 0.00/0.99
a__c → c 0.00/0.99
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2) 0.00/0.99
mark(c) → a__c 0.00/0.99
mark(b) → b
Tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1 0.00/0.99
MARK(c) → c6
S tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1
K tuples:

MARK(c) → c6
Defined Rule Symbols:

a__f, a__c, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c5, c1, c6

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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.

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1
We considered the (Usable) Rules:

mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2) 0.00/0.99
mark(c) → a__c 0.00/0.99
mark(b) → b 0.00/0.99
a__c → b 0.00/0.99
a__c → c 0.00/0.99
a__f(b, z0, c) → a__f(z0, a__c, z0) 0.00/0.99
a__f(z0, z1, z2) → f(z0, z1, z2)
And the Tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1 0.00/0.99
MARK(c) → c6
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.99

POL(A__F(x1, x2, x3)) = [4]    0.00/0.99
POL(MARK(x1)) = [2]x1    0.00/0.99
POL(a__c) = 0    0.00/0.99
POL(a__f(x1, x2, x3)) = [3] + [3]x2    0.00/0.99
POL(b) = 0    0.00/0.99
POL(c) = 0    0.00/0.99
POL(c1) = 0    0.00/0.99
POL(c5(x1, x2)) = x1 + x2    0.00/0.99
POL(c6) = 0    0.00/0.99
POL(f(x1, x2, x3)) = [4] + x1 + x2    0.00/0.99
POL(mark(x1)) = 0   
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(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(b, z0, c) → a__f(z0, a__c, z0) 0.00/0.99
a__f(z0, z1, z2) → f(z0, z1, z2) 0.00/0.99
a__c → b 0.00/0.99
a__c → c 0.00/0.99
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2) 0.00/0.99
mark(c) → a__c 0.00/0.99
mark(b) → b
Tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1 0.00/0.99
MARK(c) → c6
S tuples:none
K tuples:

MARK(c) → c6 0.00/0.99
MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1)) 0.00/0.99
A__F(b, z0, c) → c1
Defined Rule Symbols:

a__f, a__c, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c5, c1, c6

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

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

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2.43/1.03 EOF