YES(O(1), O(n^1)) 14.63/6.86 YES(O(1), O(n^1)) 14.63/6.88 14.63/6.88 14.63/6.88 14.63/6.88 14.63/6.88 14.63/6.88 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 14.63/6.88 14.63/6.88 14.63/6.88
14.63/6.88
14.63/6.88
14.63/6.88
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
14.63/6.88 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
14.63/6.88 
↳2 CdtProblem
14.63/6.88 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
14.63/6.88 
↳4 CdtProblem
14.63/6.88 
↳5 CdtInstantiationProof (BOTH BOUNDS(ID, ID))
14.63/6.88 
↳6 CdtProblem
14.63/6.88 
↳7 CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID))
14.63/6.88 
↳8 CdtProblem
14.63/6.88 
↳9 CdtLeafRemovalProof (ComplexityIfPolyImplication)
14.63/6.88 
↳10 CdtProblem
14.63/6.88 
↳11 CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
14.63/6.88 
↳12 CdtProblem
14.63/6.88 
↳13 CdtNarrowingProof (BOTH BOUNDS(ID, ID))
14.63/6.88 
↳14 CdtProblem
14.63/6.88 
↳15 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
14.63/6.88 
↳16 CdtProblem
14.63/6.88 
↳17 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
14.63/6.88 
↳18 CdtProblem
14.63/6.88 
↳19 CdtKnowledgeProof (⇔)
14.63/6.88 
↳20 BOUNDS(O(1), O(1))
14.63/6.88
14.63/6.88 14.63/6.88
14.63/6.88
14.63/6.88

(0) Obligation:

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

c(c(b(c(x)))) → b(a(0, c(x))) 14.63/6.88
c(c(x)) → b(c(b(c(x)))) 14.63/6.88
a(0, x) → c(c(x))

Rewrite Strategy: INNERMOST
14.63/6.88
14.63/6.88

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

Converted CpxTRS to CDT
14.63/6.88
14.63/6.88

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.63/6.88
c(c(z0)) → b(c(b(c(z0)))) 14.63/6.88
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.88
C(c(z0)) → c2(C(b(c(z0))), C(z0)) 14.63/6.88
A(0, z0) → c3(C(c(z0)), C(z0))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.88
C(c(z0)) → c2(C(b(c(z0))), C(z0)) 14.63/6.88
A(0, z0) → c3(C(c(z0)), C(z0))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A

Compound Symbols:

c1, c2, c3

14.63/6.88
14.63/6.88

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

Removed 1 trailing tuple parts
14.63/6.88
14.63/6.88

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.63/6.88
c(c(z0)) → b(c(b(c(z0)))) 14.63/6.88
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.88
A(0, z0) → c3(C(c(z0)), C(z0)) 14.63/6.88
C(c(z0)) → c2(C(z0))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.88
A(0, z0) → c3(C(c(z0)), C(z0)) 14.63/6.88
C(c(z0)) → c2(C(z0))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A

Compound Symbols:

c1, c3, c2

14.63/6.88
14.63/6.88

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

Use instantiation to replace A(0, z0) → c3(C(c(z0)), C(z0)) by

A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) 14.63/6.88
A1(0, z0) → c3(C(c(z0)), C(z0))
14.63/6.88
14.63/6.88

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.63/6.88
c(c(z0)) → b(c(b(c(z0)))) 14.63/6.88
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.88
C(c(z0)) → c2(C(z0)) 14.63/6.88
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) 14.63/6.88
A1(0, z0) → c3(C(c(z0)), C(z0))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.88
C(c(z0)) → c2(C(z0)) 14.63/6.88
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) 14.63/6.88
A1(0, z0) → c3(C(c(z0)), C(z0))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A, A1

Compound Symbols:

c1, c2, c3

14.63/6.89
14.63/6.89

(7) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)

Split RHS of tuples not part of any SCC
14.63/6.89
14.63/6.89

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.63/6.89
c(c(z0)) → b(c(b(c(z0)))) 14.63/6.89
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.89
C(c(z0)) → c2(C(z0)) 14.63/6.89
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) 14.63/6.89
A1(0, z0) → c4(C(c(z0))) 14.63/6.89
A1(0, z0) → c4(C(z0))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.89
C(c(z0)) → c2(C(z0)) 14.63/6.89
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) 14.63/6.89
A1(0, z0) → c4(C(c(z0))) 14.63/6.89
A1(0, z0) → c4(C(z0))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A, A1

Compound Symbols:

c1, c2, c3, c4

14.63/6.89
14.63/6.89

(9) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

A1(0, z0) → c4(C(z0))
14.63/6.89
14.63/6.89

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.63/6.89
c(c(z0)) → b(c(b(c(z0)))) 14.63/6.89
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.89
C(c(z0)) → c2(C(z0)) 14.63/6.89
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) 14.63/6.89
A1(0, z0) → c4(C(c(z0)))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.89
C(c(z0)) → c2(C(z0)) 14.63/6.89
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) 14.63/6.89
A1(0, z0) → c4(C(c(z0)))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A, A1

Compound Symbols:

c1, c2, c3, c4

14.63/6.89
14.63/6.89

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

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

A1(0, z0) → c4(C(c(z0)))
14.63/6.89
14.63/6.89

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.63/6.89
c(c(z0)) → b(c(b(c(z0)))) 14.63/6.89
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.89
C(c(z0)) → c2(C(z0)) 14.63/6.89
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) 14.63/6.89
A1(0, z0) → c4(C(c(z0)))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.63/6.89
C(c(z0)) → c2(C(z0)) 14.63/6.89
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
K tuples:

A1(0, z0) → c4(C(c(z0)))
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A, A1

Compound Symbols:

c1, c2, c3, c4

14.63/6.89
14.63/6.89

(13) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) by

A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0))))) 14.63/6.89
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
14.63/6.89
14.63/6.89

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.63/6.89
c(c(z0)) → b(c(b(c(z0)))) 14.63/6.89
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0)) 14.98/6.90
A1(0, z0) → c4(C(c(z0))) 14.98/6.90
A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0))))) 14.98/6.90
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0)) 14.98/6.90
A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0))))) 14.98/6.90
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
K tuples:

A1(0, z0) → c4(C(c(z0)))
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A1, A

Compound Symbols:

c1, c2, c4, c3

14.98/6.90
14.98/6.90

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

Removed 2 trailing tuple parts
14.98/6.90
14.98/6.90

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.98/6.90
c(c(z0)) → b(c(b(c(z0)))) 14.98/6.90
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0)) 14.98/6.90
A1(0, z0) → c4(C(c(z0))) 14.98/6.90
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0))))) 14.98/6.90
A(0, c(z0)) → c3(C(c(z0)))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0)) 14.98/6.90
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0))))) 14.98/6.90
A(0, c(z0)) → c3(C(c(z0)))
K tuples:

A1(0, z0) → c4(C(c(z0)))
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A1, A

Compound Symbols:

c1, c2, c4, c3

14.98/6.90
14.98/6.90

(17) 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.

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0))
We considered the (Usable) Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.98/6.90
c(c(z0)) → b(c(b(c(z0)))) 14.98/6.90
a(0, z0) → c(c(z0))
And the Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0)) 14.98/6.90
A1(0, z0) → c4(C(c(z0))) 14.98/6.90
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0))))) 14.98/6.90
A(0, c(z0)) → c3(C(c(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 14.98/6.90

POL(0) = [3]    14.98/6.90
POL(A(x1, x2)) = x2    14.98/6.90
POL(A1(x1, x2)) = [1] + x1 + [5]x2    14.98/6.90
POL(C(x1)) = x1    14.98/6.90
POL(a(x1, x2)) = [3] + [3]x1 + [4]x2    14.98/6.90
POL(b(x1)) = x1    14.98/6.90
POL(c(x1)) = [4] + [2]x1    14.98/6.90
POL(c1(x1, x2)) = x1 + x2    14.98/6.90
POL(c2(x1)) = x1    14.98/6.90
POL(c3(x1)) = x1    14.98/6.90
POL(c4(x1)) = x1   
14.98/6.90
14.98/6.90

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0))) 14.98/6.90
c(c(z0)) → b(c(b(c(z0)))) 14.98/6.90
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0)) 14.98/6.90
A1(0, z0) → c4(C(c(z0))) 14.98/6.90
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0))))) 14.98/6.90
A(0, c(z0)) → c3(C(c(z0)))
S tuples:

A(0, c(b(c(z0)))) → c3(C(c(b(c(z0))))) 14.98/6.90
A(0, c(z0)) → c3(C(c(z0)))
K tuples:

A1(0, z0) → c4(C(c(z0))) 14.98/6.90
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0))
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A1, A

Compound Symbols:

c1, c2, c4, c3

14.98/6.90
14.98/6.90

(19) CdtKnowledgeProof (EQUIVALENT transformation)

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

A(0, c(b(c(z0)))) → c3(C(c(b(c(z0))))) 14.98/6.90
A(0, c(z0)) → c3(C(c(z0))) 14.98/6.90
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0)) 14.98/6.90
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0)) 14.98/6.90
C(c(z0)) → c2(C(z0))
Now S is empty
14.98/6.90
14.98/6.90

(20) BOUNDS(O(1), O(1))

14.98/6.90
14.98/6.90
14.98/6.97 EOF