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:

f(f(x)) → f(g(f(x), x))
f(f(x)) → f(h(f(x), f(x)))
g(x, y) → y
h(x, x) → g(x, 0)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(g(f(z0), z0))
f(f(z0)) → f(h(f(z0), f(z0)))
g(z0, z1) → z1
h(z0, z0) → g(z0, 0)

Tuples:

F(f(z0)) → c(F(g(f(z0), z0)), G(f(z0), z0), F(z0))
F(f(z0)) → c1(F(h(f(z0), f(z0))), H(f(z0), f(z0)), F(z0), F(z0))
H(z0, z0) → c3(G(z0, 0))

S tuples:

F(f(z0)) → c(F(g(f(z0), z0)), G(f(z0), z0), F(z0))
F(f(z0)) → c1(F(h(f(z0), f(z0))), H(f(z0), f(z0)), F(z0), F(z0))
H(z0, z0) → c3(G(z0, 0))

K tuples:none
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F, H

Compound Symbols:

c, c1, c3

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

H(z0, z0) → c3(G(z0, 0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(g(f(z0), z0))
f(f(z0)) → f(h(f(z0), f(z0)))
g(z0, z1) → z1
h(z0, z0) → g(z0, 0)

Tuples:

F(f(z0)) → c(F(g(f(z0), z0)), G(f(z0), z0), F(z0))
F(f(z0)) → c1(F(h(f(z0), f(z0))), H(f(z0), f(z0)), F(z0), F(z0))

S tuples:

F(f(z0)) → c(F(g(f(z0), z0)), G(f(z0), z0), F(z0))
F(f(z0)) → c1(F(h(f(z0), f(z0))), H(f(z0), f(z0)), F(z0), F(z0))

K tuples:none
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c, 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.

F(f(z0)) → c(F(g(f(z0), z0)), G(f(z0), z0), F(z0))
F(f(z0)) → c1(F(h(f(z0), f(z0))), H(f(z0), f(z0)), F(z0), F(z0))

We considered the (Usable) Rules:

h(z0, z0) → g(z0, 0)
g(z0, z1) → z1

And the Tuples:

F(f(z0)) → c(F(g(f(z0), z0)), G(f(z0), z0), F(z0))
F(f(z0)) → c1(F(h(f(z0), f(z0))), H(f(z0), f(z0)), F(z0), F(z0))

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

POL(0) = 0   
POL(F(x₁)) = x₁   
POL(G(x₁, x₂)) = 0   
POL(H(x₁, x₂)) = 0   
POL(c(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c1(x₁, x₂, x₃, x₄)) = x₁ + x₂ + x₃ + x₄   
POL(f(x₁)) = [2] + [2]x₁   
POL(g(x₁, x₂)) = x₂   
POL(h(x₁, x₂)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(g(f(z0), z0))
f(f(z0)) → f(h(f(z0), f(z0)))
g(z0, z1) → z1
h(z0, z0) → g(z0, 0)

Tuples:

F(f(z0)) → c(F(g(f(z0), z0)), G(f(z0), z0), F(z0))
F(f(z0)) → c1(F(h(f(z0), f(z0))), H(f(z0), f(z0)), F(z0), F(z0))

S tuples:none
K tuples:

F(f(z0)) → c(F(g(f(z0), z0)), G(f(z0), z0), F(z0))
F(f(z0)) → c1(F(h(f(z0), f(z0))), H(f(z0), f(z0)), F(z0), F(z0))

Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c, c1

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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