The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

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

f(f(X, Y), Z) → f(X, f(Y, Z))
f(X, f(Y, Z)) → f(Y, Y)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0, z1), z2) → f(z0, f(z1, z2))
f(z0, f(z1, z2)) → f(z1, z1)

Tuples:

F(f(z0, z1), z2) → c(F(z0, f(z1, z2)), F(z1, z2))
F(z0, f(z1, z2)) → c1(F(z1, z1))

S tuples:

F(f(z0, z1), z2) → c(F(z0, f(z1, z2)), F(z1, z2))
F(z0, f(z1, z2)) → c1(F(z1, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

(3) 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.

F(f(z0, z1), z2) → c(F(z0, f(z1, z2)), F(z1, z2))
F(z0, f(z1, z2)) → c1(F(z1, z1))

We considered the (Usable) Rules:

f(z0, f(z1, z2)) → f(z1, z1)

And the Tuples:

F(f(z0, z1), z2) → c(F(z0, f(z1, z2)), F(z1, z2))
F(z0, f(z1, z2)) → c1(F(z1, z1))

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

POL(F(x₁, x₂)) = [2]x₁ + [2]x₂ + [2]x₁·x₂ + [2]x₁²   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁)) = x₁   
POL(f(x₁, x₂)) = [2] + [2]x₁ + [2]x₂ + x₁·x₂ + [2]x₁²   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0, z1), z2) → f(z0, f(z1, z2))
f(z0, f(z1, z2)) → f(z1, z1)

Tuples:

F(f(z0, z1), z2) → c(F(z0, f(z1, z2)), F(z1, z2))
F(z0, f(z1, z2)) → c1(F(z1, z1))

S tuples:none
K tuples:

F(f(z0, z1), z2) → c(F(z0, f(z1, z2)), F(z1, z2))
F(z0, f(z1, z2)) → c1(F(z1, z1))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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