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(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(c(z0))) → c(c(a(z0, 0)))
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
c(z0) → z0

Tuples:

C(c(c(z0))) → c1(C(c(a(z0, 0))), C(a(z0, 0)))
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))

S tuples:

C(c(c(z0))) → c1(C(c(a(z0, 0))), C(a(z0, 0)))
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))

K tuples:none
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c1, c2

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

C(c(c(z0))) → c1(C(c(a(z0, 0))), C(a(z0, 0)))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(c(z0))) → c(c(a(z0, 0)))
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
c(z0) → z0

Tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))

S tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))

K tuples:none
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c2

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(c(z0))) → c(c(a(z0, 0)))
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
c(z0) → z0

Tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))

S tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))

K tuples:none
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c2

(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(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))

We considered the (Usable) Rules:

c(z0) → z0
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)

And the Tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))

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

POL(0) = 0   
POL(C(x₁)) = [2]x₁   
POL(a(x₁, x₂)) = [1] + x₂   
POL(c(x₁)) = [2]x₁   
POL(c2(x₁, x₂)) = x₁ + x₂   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(c(z0))) → c(c(a(z0, 0)))
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
c(z0) → z0

Tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))

S tuples:none
K tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))

Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c2

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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