(0) Obligation:

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

f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(g(z0), z0, z1) → c(F(z1, z1, g(z1)), G(z1))
G(g(z0)) → c1(G(z0))
S tuples:

F(g(z0), z0, z1) → c(F(z1, z1, g(z1)), G(z1))
G(g(z0)) → c1(G(z0))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

F(g(z0), z0, z1) → c(F(z1, z1, g(z1)), G(z1))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

G(g(z0)) → c1(G(z0))
S tuples:

G(g(z0)) → c1(G(z0))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

G

Compound Symbols:

c1

(5) 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.

G(g(z0)) → c1(G(z0))
We considered the (Usable) Rules:none
And the Tuples:

G(g(z0)) → c1(G(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(G(x1)) = x1   
POL(c1(x1)) = x1   
POL(g(x1)) = [1] + [4]x1   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

G(g(z0)) → c1(G(z0))
S tuples:none
K tuples:

G(g(z0)) → c1(G(z0))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

G

Compound Symbols:

c1

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))