The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

(0) Obligation:

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

c(a(b(a(b(x1))))) → a(b(a(b(b(a(b(b(c(a(b(c(a(x1)))))))))))))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(a(b(a(b(z0))))) → a(b(a(b(b(a(b(b(c(a(b(c(a(z0)))))))))))))

Tuples:

C(a(b(a(b(z0))))) → c1(C(a(b(c(a(z0))))), C(a(z0)))

S tuples:

C(a(b(a(b(z0))))) → c1(C(a(b(c(a(z0))))), C(a(z0)))

K tuples:none
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c1

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

Use narrowing to replace C(a(b(a(b(z0))))) → c1(C(a(b(c(a(z0))))), C(a(z0))) by

C(a(b(a(b(b(a(b(z0)))))))) → c1(C(a(b(a(b(a(b(b(a(b(b(c(a(b(c(a(z0)))))))))))))))), C(a(b(a(b(z0))))))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(a(b(a(b(z0))))) → a(b(a(b(b(a(b(b(c(a(b(c(a(z0)))))))))))))

Tuples:

C(a(b(a(b(b(a(b(z0)))))))) → c1(C(a(b(a(b(a(b(b(a(b(b(c(a(b(c(a(z0)))))))))))))))), C(a(b(a(b(z0))))))

S tuples:

C(a(b(a(b(b(a(b(z0)))))))) → c1(C(a(b(a(b(a(b(b(a(b(b(c(a(b(c(a(z0)))))))))))))))), C(a(b(a(b(z0))))))

K tuples:none
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c1

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(a(b(a(b(z0))))) → a(b(a(b(b(a(b(b(c(a(b(c(a(z0)))))))))))))

Tuples:

C(a(b(a(b(b(a(b(z0)))))))) → c1(C(a(b(a(b(z0))))))

S tuples:

C(a(b(a(b(b(a(b(z0)))))))) → c1(C(a(b(a(b(z0))))))

K tuples:none
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c1

(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.

C(a(b(a(b(b(a(b(z0)))))))) → c1(C(a(b(a(b(z0))))))

We considered the (Usable) Rules:none
And the Tuples:

C(a(b(a(b(b(a(b(z0)))))))) → c1(C(a(b(a(b(z0))))))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(C(x₁)) = [2]x₁   
POL(a(x₁)) = [1] + x₁   
POL(b(x₁)) = x₁   
POL(c1(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(a(b(a(b(z0))))) → a(b(a(b(b(a(b(b(c(a(b(c(a(z0)))))))))))))

Tuples:

C(a(b(a(b(b(a(b(z0)))))))) → c1(C(a(b(a(b(z0))))))

S tuples:none
K tuples:

C(a(b(a(b(b(a(b(z0)))))))) → c1(C(a(b(a(b(z0))))))

Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c1

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))