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:

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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))

S tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))

K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

(3) 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(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))

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

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))

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

POL(G(x₁, x₂)) = [4]x₁ + [5]x₂   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(f(x₁, x₂)) = [2] + x₂   
POL(h(x₁, x₂)) = [5] + x₁ + x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))

S tuples:

G(z0, h(z1, z2)) → c2(G(z0, z1))

K tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))

Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

(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(z0, h(z1, z2)) → c2(G(z0, z1))

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

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))

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

POL(G(x₁, x₂)) = [4]x₁ + x₂   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(f(x₁, x₂)) = [3] + x₁ + x₂   
POL(h(x₁, x₂)) = [1] + x₁ + x₂   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))

S tuples:none
K tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))

Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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