YES(O(1), O(n^1)) 2.50/1.04 YES(O(1), O(n^1)) 2.50/1.06 2.50/1.06 2.50/1.06 2.50/1.06 2.50/1.06 2.50/1.06 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 2.50/1.06 2.50/1.06 2.50/1.06
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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))
2.50/1.06 
↳4 CdtProblem
2.50/1.06 
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
2.50/1.06 
↳6 CdtProblem
2.50/1.06 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
2.50/1.06 
↳8 CdtProblem
2.50/1.06 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
2.50/1.06 
↳10 CdtProblem
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↳11 CdtKnowledgeProof (⇔)
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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(f(X)) → a__c(f(g(f(X)))) 2.50/1.06
a__c(X) → d(X) 2.50/1.06
a__h(X) → a__c(d(X)) 2.50/1.06
mark(f(X)) → a__f(mark(X)) 2.50/1.06
mark(c(X)) → a__c(X) 2.50/1.06
mark(h(X)) → a__h(mark(X)) 2.50/1.06
mark(g(X)) → g(X) 2.50/1.06
mark(d(X)) → d(X) 2.50/1.06
a__f(X) → f(X) 2.50/1.06
a__c(X) → c(X) 2.50/1.06
a__h(X) → h(X)

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

Converted CpxTRS to CDT
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2.50/1.06

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(z0)) → a__c(f(g(f(z0)))) 2.50/1.06
a__f(z0) → f(z0) 2.50/1.06
a__c(z0) → d(z0) 2.50/1.06
a__c(z0) → c(z0) 2.50/1.06
a__h(z0) → a__c(d(z0)) 2.50/1.06
a__h(z0) → h(z0) 2.50/1.06
mark(f(z0)) → a__f(mark(z0)) 2.50/1.06
mark(c(z0)) → a__c(z0) 2.50/1.06
mark(h(z0)) → a__h(mark(z0)) 2.50/1.06
mark(g(z0)) → g(z0) 2.50/1.06
mark(d(z0)) → d(z0)
Tuples:

A__F(f(z0)) → c1(A__C(f(g(f(z0))))) 2.50/1.06
A__H(z0) → c5(A__C(d(z0))) 2.50/1.06
MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.06
MARK(c(z0)) → c8(A__C(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0))
S tuples:

A__F(f(z0)) → c1(A__C(f(g(f(z0))))) 2.50/1.07
A__H(z0) → c5(A__C(d(z0))) 2.50/1.07
MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(c(z0)) → c8(A__C(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, a__c, a__h, mark

Defined Pair Symbols:

A__F, A__H, MARK

Compound Symbols:

c1, c5, c7, c8, c9

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2.50/1.07

(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(f(z0)) → a__c(f(g(f(z0)))) 2.50/1.07
a__f(z0) → f(z0) 2.50/1.07
a__c(z0) → d(z0) 2.50/1.07
a__c(z0) → c(z0) 2.50/1.07
a__h(z0) → a__c(d(z0)) 2.50/1.07
a__h(z0) → h(z0) 2.50/1.07
mark(f(z0)) → a__f(mark(z0)) 2.50/1.07
mark(c(z0)) → a__c(z0) 2.50/1.07
mark(h(z0)) → a__h(mark(z0)) 2.50/1.07
mark(g(z0)) → g(z0) 2.50/1.07
mark(d(z0)) → d(z0)
Tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
A__H(z0) → c5 2.50/1.07
MARK(c(z0)) → c8
S tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
A__H(z0) → c5 2.50/1.07
MARK(c(z0)) → c8
K tuples:none
Defined Rule Symbols:

a__f, a__c, a__h, mark

Defined Pair Symbols:

MARK, A__F, A__H

Compound Symbols:

c7, c9, c1, c5, c8

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

Removed 3 trailing nodes:

A__H(z0) → c5 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
MARK(c(z0)) → c8
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(z0)) → a__c(f(g(f(z0)))) 2.50/1.07
a__f(z0) → f(z0) 2.50/1.07
a__c(z0) → d(z0) 2.50/1.07
a__c(z0) → c(z0) 2.50/1.07
a__h(z0) → a__c(d(z0)) 2.50/1.07
a__h(z0) → h(z0) 2.50/1.07
mark(f(z0)) → a__f(mark(z0)) 2.50/1.07
mark(c(z0)) → a__c(z0) 2.50/1.07
mark(h(z0)) → a__h(mark(z0)) 2.50/1.07
mark(g(z0)) → g(z0) 2.50/1.07
mark(d(z0)) → d(z0)
Tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
A__H(z0) → c5 2.50/1.07
MARK(c(z0)) → c8
S tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
A__H(z0) → c5 2.50/1.07
MARK(c(z0)) → c8
K tuples:none
Defined Rule Symbols:

a__f, a__c, a__h, mark

Defined Pair Symbols:

MARK, A__F, A__H

Compound Symbols:

c7, c9, c1, c5, c8

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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(z0)) → c8
We considered the (Usable) Rules:

mark(f(z0)) → a__f(mark(z0)) 2.50/1.07
mark(c(z0)) → a__c(z0) 2.50/1.07
mark(h(z0)) → a__h(mark(z0)) 2.50/1.07
mark(g(z0)) → g(z0) 2.50/1.07
mark(d(z0)) → d(z0) 2.50/1.07
a__h(z0) → a__c(d(z0)) 2.50/1.07
a__h(z0) → h(z0) 2.50/1.07
a__c(z0) → d(z0) 2.50/1.07
a__c(z0) → c(z0) 2.50/1.07
a__f(f(z0)) → a__c(f(g(f(z0)))) 2.50/1.07
a__f(z0) → f(z0)
And the Tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
A__H(z0) → c5 2.50/1.07
MARK(c(z0)) → c8
The order we found is given by the following interpretation:
Polynomial interpretation : 2.50/1.07

POL(A__F(x1)) = 0    2.50/1.07
POL(A__H(x1)) = 0    2.50/1.07
POL(MARK(x1)) = [3]    2.50/1.07
POL(a__c(x1)) = x1    2.50/1.07
POL(a__f(x1)) = [5] + x1    2.50/1.07
POL(a__h(x1)) = [4]    2.50/1.07
POL(c(x1)) = x1    2.50/1.07
POL(c1) = 0    2.50/1.07
POL(c5) = 0    2.50/1.07
POL(c7(x1, x2)) = x1 + x2    2.50/1.07
POL(c8) = 0    2.50/1.07
POL(c9(x1, x2)) = x1 + x2    2.50/1.07
POL(d(x1)) = 0    2.50/1.07
POL(f(x1)) = [3] + x1    2.50/1.07
POL(g(x1)) = [5]    2.50/1.07
POL(h(x1)) = [2]    2.50/1.07
POL(mark(x1)) = [3]x1   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(z0)) → a__c(f(g(f(z0)))) 2.50/1.07
a__f(z0) → f(z0) 2.50/1.07
a__c(z0) → d(z0) 2.50/1.07
a__c(z0) → c(z0) 2.50/1.07
a__h(z0) → a__c(d(z0)) 2.50/1.07
a__h(z0) → h(z0) 2.50/1.07
mark(f(z0)) → a__f(mark(z0)) 2.50/1.07
mark(c(z0)) → a__c(z0) 2.50/1.07
mark(h(z0)) → a__h(mark(z0)) 2.50/1.07
mark(g(z0)) → g(z0) 2.50/1.07
mark(d(z0)) → d(z0)
Tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
A__H(z0) → c5 2.50/1.07
MARK(c(z0)) → c8
S tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
A__H(z0) → c5
K tuples:

MARK(c(z0)) → c8
Defined Rule Symbols:

a__f, a__c, a__h, mark

Defined Pair Symbols:

MARK, A__F, A__H

Compound Symbols:

c7, c9, c1, c5, c8

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2.50/1.07

(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)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__H(z0) → c5
We considered the (Usable) Rules:

mark(f(z0)) → a__f(mark(z0)) 2.50/1.07
mark(c(z0)) → a__c(z0) 2.50/1.07
mark(h(z0)) → a__h(mark(z0)) 2.50/1.07
mark(g(z0)) → g(z0) 2.50/1.07
mark(d(z0)) → d(z0) 2.50/1.07
a__h(z0) → a__c(d(z0)) 2.50/1.07
a__h(z0) → h(z0) 2.50/1.07
a__c(z0) → d(z0) 2.50/1.07
a__c(z0) → c(z0) 2.50/1.07
a__f(f(z0)) → a__c(f(g(f(z0)))) 2.50/1.07
a__f(z0) → f(z0)
And the Tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
A__H(z0) → c5 2.50/1.07
MARK(c(z0)) → c8
The order we found is given by the following interpretation:
Polynomial interpretation : 2.50/1.07

POL(A__F(x1)) = 0    2.50/1.07
POL(A__H(x1)) = [3]    2.50/1.07
POL(MARK(x1)) = [2]x1    2.50/1.07
POL(a__c(x1)) = x1    2.50/1.07
POL(a__f(x1)) = [5] + x1    2.50/1.07
POL(a__h(x1)) = [2] + x1    2.50/1.07
POL(c(x1)) = x1    2.50/1.07
POL(c1) = 0    2.50/1.07
POL(c5) = 0    2.50/1.07
POL(c7(x1, x2)) = x1 + x2    2.50/1.07
POL(c8) = 0    2.50/1.07
POL(c9(x1, x2)) = x1 + x2    2.50/1.07
POL(d(x1)) = 0    2.50/1.07
POL(f(x1)) = [3] + x1    2.50/1.07
POL(g(x1)) = [5]    2.50/1.07
POL(h(x1)) = [2] + x1    2.50/1.07
POL(mark(x1)) = [2] + [5]x1   
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2.50/1.07

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(z0)) → a__c(f(g(f(z0)))) 2.50/1.07
a__f(z0) → f(z0) 2.50/1.07
a__c(z0) → d(z0) 2.50/1.07
a__c(z0) → c(z0) 2.50/1.07
a__h(z0) → a__c(d(z0)) 2.50/1.07
a__h(z0) → h(z0) 2.50/1.07
mark(f(z0)) → a__f(mark(z0)) 2.50/1.07
mark(c(z0)) → a__c(z0) 2.50/1.07
mark(h(z0)) → a__h(mark(z0)) 2.50/1.07
mark(g(z0)) → g(z0) 2.50/1.07
mark(d(z0)) → d(z0)
Tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__F(f(z0)) → c1 2.50/1.07
A__H(z0) → c5 2.50/1.07
MARK(c(z0)) → c8
S tuples:

A__F(f(z0)) → c1
K tuples:

MARK(c(z0)) → c8 2.50/1.07
MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0)) 2.50/1.07
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0)) 2.50/1.07
A__H(z0) → c5
Defined Rule Symbols:

a__f, a__c, a__h, mark

Defined Pair Symbols:

MARK, A__F, A__H

Compound Symbols:

c7, c9, c1, c5, c8

2.50/1.07
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(11) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

A__F(f(z0)) → c1
Now S is empty
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(12) BOUNDS(O(1), O(1))

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2.82/1.12 EOF