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:

.(.(x, y), z) → .(x, .(y, z))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

.(.(z0, z1), z2) → .(z0, .(z1, z2))

Tuples:

.'(.(z0, z1), z2) → c(.'(z0, .(z1, z2)), .'(z1, z2))

S tuples:

.'(.(z0, z1), z2) → c(.'(z0, .(z1, z2)), .'(z1, z2))

K tuples:none
Defined Rule Symbols:

.

Defined Pair Symbols:

.'

Compound Symbols:

c

(3) 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.

.'(.(z0, z1), z2) → c(.'(z0, .(z1, z2)), .'(z1, z2))

We considered the (Usable) Rules:

.(.(z0, z1), z2) → .(z0, .(z1, z2))

And the Tuples:

.'(.(z0, z1), z2) → c(.'(z0, .(z1, z2)), .'(z1, z2))

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

POL(.(x₁, x₂)) = [4] + [4]x₁ + [4]x₂   
POL(.'(x₁, x₂)) = [1] + [4]x₁   
POL(c(x₁, x₂)) = x₁ + x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

.(.(z0, z1), z2) → .(z0, .(z1, z2))

Tuples:

.'(.(z0, z1), z2) → c(.'(z0, .(z1, z2)), .'(z1, z2))

S tuples:none
K tuples:

.'(.(z0, z1), z2) → c(.'(z0, .(z1, z2)), .'(z1, z2))

Defined Rule Symbols:

.

Defined Pair Symbols:

.'

Compound Symbols:

c

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(6) BOUNDS(O(1), O(1))