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:

h(f(x), y) → f(g(x, y))
g(x, y) → h(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:

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

Tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))

S tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))

K tuples:none
Defined Rule Symbols:

h, g

Defined Pair Symbols:

H, G

Compound Symbols:

c, c1

(3) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))

S tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

H, G

Compound Symbols:

c, c1

(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.

H(f(z0), z1) → c(G(z0, z1))

We considered the (Usable) Rules:none
And the Tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))

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

POL(G(x₁, x₂)) = [2]x₁   
POL(H(x₁, x₂)) = [2]x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(f(x₁)) = [4] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))

S tuples:

G(z0, z1) → c1(H(z0, z1))

K tuples:

H(f(z0), z1) → c(G(z0, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

H, G

Compound Symbols:

c, c1

(7) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

G(z0, z1) → c1(H(z0, z1))
H(f(z0), z1) → c(G(z0, z1))

Now S is empty

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