### (0) Obligation:

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

f(g(x), y, y) → g(f(x, x, y))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(g(z0), z1, z1) → g(f(z0, z0, z1))
Tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))
S tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

### (3) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(g(z0), z1, z1) → g(f(z0, z0, z1))

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))
S tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

### (5) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F(g(z0), z1, z1) → c(F(z0, z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2, x3)) = [2]x1
POL(c(x1)) = x1
POL(g(x1)) = [1] + x1

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))
S tuples:none
K tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

### (7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty