YES(O(1), O(n^2)) 0.00/0.94 YES(O(1), O(n^2)) 0.00/0.96 0.00/0.96 0.00/0.96 0.00/0.96 0.00/0.96 0.00/0.96 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.96 0.00/0.96 0.00/0.96
0.00/0.96
0.00/0.96
0.00/0.96
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
0.00/0.96 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/0.96 
↳2 CdtProblem
0.00/0.96 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
0.00/0.96 
↳4 CdtProblem
0.00/0.96 
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
0.00/0.96 
↳6 CdtProblem
0.00/0.96 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.96 
↳8 CdtProblem
0.00/0.96 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
0.00/0.96 
↳10 CdtProblem
0.00/0.96 
↳11 CdtKnowledgeProof (⇔)
0.00/0.96 
↳12 BOUNDS(O(1), O(1))
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0.00/0.96 0.00/0.96
0.00/0.96
0.00/0.96

(0) Obligation:

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

a__f(X, X) → a__f(a, b) 0.00/0.96
a__b → a 0.00/0.96
mark(f(X1, X2)) → a__f(mark(X1), X2) 0.00/0.96
mark(b) → a__b 0.00/0.96
mark(a) → a 0.00/0.96
a__f(X1, X2) → f(X1, X2) 0.00/0.96
a__b → b

Rewrite Strategy: INNERMOST
0.00/0.96
0.00/0.96

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
0.00/0.96
0.00/0.96

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, z0) → a__f(a, b) 0.00/0.96
a__f(z0, z1) → f(z0, z1) 0.00/0.96
a__b → a 0.00/0.96
a__b → b 0.00/0.96
mark(f(z0, z1)) → a__f(mark(z0), z1) 0.00/0.96
mark(b) → a__b 0.00/0.96
mark(a) → a
Tuples:

A__F(z0, z0) → c(A__F(a, b)) 0.00/0.96
MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
MARK(b) → c5(A__B)
S tuples:

A__F(z0, z0) → c(A__F(a, b)) 0.00/0.96
MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
MARK(b) → c5(A__B)
K tuples:none
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c, c4, c5

0.00/0.96
0.00/0.96

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

Removed 2 trailing tuple parts
0.00/0.96
0.00/0.96

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, z0) → a__f(a, b) 0.00/0.96
a__f(z0, z1) → f(z0, z1) 0.00/0.96
a__b → a 0.00/0.96
a__b → b 0.00/0.96
mark(f(z0, z1)) → a__f(mark(z0), z1) 0.00/0.96
mark(b) → a__b 0.00/0.96
mark(a) → a
Tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
A__F(z0, z0) → c 0.00/0.96
MARK(b) → c5
S tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
A__F(z0, z0) → c 0.00/0.96
MARK(b) → c5
K tuples:none
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c4, c, c5

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0.00/0.96

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

Removed 2 trailing nodes:

MARK(b) → c5 0.00/0.96
A__F(z0, z0) → c
0.00/0.96
0.00/0.96

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, z0) → a__f(a, b) 0.00/0.96
a__f(z0, z1) → f(z0, z1) 0.00/0.96
a__b → a 0.00/0.96
a__b → b 0.00/0.96
mark(f(z0, z1)) → a__f(mark(z0), z1) 0.00/0.96
mark(b) → a__b 0.00/0.96
mark(a) → a
Tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
A__F(z0, z0) → c 0.00/0.96
MARK(b) → c5
S tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
A__F(z0, z0) → c 0.00/0.96
MARK(b) → c5
K tuples:none
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c4, c, c5

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0.00/0.96

(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(b) → c5
We considered the (Usable) Rules:

mark(f(z0, z1)) → a__f(mark(z0), z1) 0.00/0.96
mark(b) → a__b 0.00/0.96
mark(a) → a 0.00/0.96
a__b → a 0.00/0.96
a__b → b 0.00/0.96
a__f(z0, z0) → a__f(a, b) 0.00/0.96
a__f(z0, z1) → f(z0, z1)
And the Tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
A__F(z0, z0) → c 0.00/0.96
MARK(b) → c5
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.96

POL(A__F(x1, x2)) = 0    0.00/0.96
POL(MARK(x1)) = [5]    0.00/0.96
POL(a) = [2]    0.00/0.96
POL(a__b) = [4]    0.00/0.96
POL(a__f(x1, x2)) = [4]    0.00/0.96
POL(b) = 0    0.00/0.96
POL(c) = 0    0.00/0.96
POL(c4(x1, x2)) = x1 + x2    0.00/0.96
POL(c5) = 0    0.00/0.96
POL(f(x1, x2)) = 0    0.00/0.96
POL(mark(x1)) = [4]   
0.00/0.96
0.00/0.96

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, z0) → a__f(a, b) 0.00/0.96
a__f(z0, z1) → f(z0, z1) 0.00/0.96
a__b → a 0.00/0.96
a__b → b 0.00/0.96
mark(f(z0, z1)) → a__f(mark(z0), z1) 0.00/0.96
mark(b) → a__b 0.00/0.96
mark(a) → a
Tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
A__F(z0, z0) → c 0.00/0.96
MARK(b) → c5
S tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
A__F(z0, z0) → c
K tuples:

MARK(b) → c5
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c4, c, c5

0.00/0.96
0.00/0.96

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0))
We considered the (Usable) Rules:

mark(f(z0, z1)) → a__f(mark(z0), z1) 0.00/0.96
mark(b) → a__b 0.00/0.96
mark(a) → a 0.00/0.96
a__b → a 0.00/0.96
a__b → b 0.00/0.96
a__f(z0, z0) → a__f(a, b) 0.00/0.96
a__f(z0, z1) → f(z0, z1)
And the Tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
A__F(z0, z0) → c 0.00/0.96
MARK(b) → c5
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.96

POL(A__F(x1, x2)) = 0    0.00/0.96
POL(MARK(x1)) = x12    0.00/0.96
POL(a) = 0    0.00/0.96
POL(a__b) = 0    0.00/0.96
POL(a__f(x1, x2)) = [2] + x1    0.00/0.96
POL(b) = 0    0.00/0.96
POL(c) = 0    0.00/0.96
POL(c4(x1, x2)) = x1 + x2    0.00/0.96
POL(c5) = 0    0.00/0.96
POL(f(x1, x2)) = [1] + x1    0.00/0.96
POL(mark(x1)) = [3]x12   
0.00/0.96
0.00/0.96

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, z0) → a__f(a, b) 0.00/0.96
a__f(z0, z1) → f(z0, z1) 0.00/0.96
a__b → a 0.00/0.96
a__b → b 0.00/0.96
mark(f(z0, z1)) → a__f(mark(z0), z1) 0.00/0.96
mark(b) → a__b 0.00/0.96
mark(a) → a
Tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0)) 0.00/0.96
A__F(z0, z0) → c 0.00/0.96
MARK(b) → c5
S tuples:

A__F(z0, z0) → c
K tuples:

MARK(b) → c5 0.00/0.96
MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0))
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c4, c, c5

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0.00/0.96

(11) CdtKnowledgeProof (EQUIVALENT transformation)

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

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

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0.00/1.00 EOF