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(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(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(f(z0)) → f(c(f(z0)))
f(f(z0)) → f(d(f(z0)))
g(c(z0)) → z0
g(d(z0)) → z0
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(z0)) → g(z0)

Tuples:

F(f(z0)) → c1(F(c(f(z0))), F(z0))
F(f(z0)) → c2(F(d(f(z0))), F(z0))
G(c(h(0))) → c5(G(d(1)))
G(c(1)) → c6(G(d(h(0))))
G(h(z0)) → c7(G(z0))

S tuples:

F(f(z0)) → c1(F(c(f(z0))), F(z0))
F(f(z0)) → c2(F(d(f(z0))), F(z0))
G(c(h(0))) → c5(G(d(1)))
G(c(1)) → c6(G(d(h(0))))
G(h(z0)) → c7(G(z0))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2, c5, c6, c7

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

Removed 2 trailing nodes:

G(c(1)) → c6(G(d(h(0))))
G(c(h(0))) → c5(G(d(1)))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(c(f(z0)))
f(f(z0)) → f(d(f(z0)))
g(c(z0)) → z0
g(d(z0)) → z0
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(z0)) → g(z0)

Tuples:

F(f(z0)) → c1(F(c(f(z0))), F(z0))
F(f(z0)) → c2(F(d(f(z0))), F(z0))
G(h(z0)) → c7(G(z0))

S tuples:

F(f(z0)) → c1(F(c(f(z0))), F(z0))
F(f(z0)) → c2(F(d(f(z0))), F(z0))
G(h(z0)) → c7(G(z0))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2, c7

(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)) → c1(F(c(f(z0))), F(z0))
F(f(z0)) → c2(F(d(f(z0))), F(z0))
G(h(z0)) → c7(G(z0))

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

F(f(z0)) → c1(F(c(f(z0))), F(z0))
F(f(z0)) → c2(F(d(f(z0))), F(z0))
G(h(z0)) → c7(G(z0))

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

POL(F(x₁)) = [5]x₁   
POL(G(x₁)) = [5]x₁   
POL(c(x₁)) = 0   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(d(x₁)) = 0   
POL(f(x₁)) = [4] + [2]x₁   
POL(h(x₁)) = [5] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(c(f(z0)))
f(f(z0)) → f(d(f(z0)))
g(c(z0)) → z0
g(d(z0)) → z0
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(z0)) → g(z0)

Tuples:

F(f(z0)) → c1(F(c(f(z0))), F(z0))
F(f(z0)) → c2(F(d(f(z0))), F(z0))
G(h(z0)) → c7(G(z0))

S tuples:none
K tuples:

F(f(z0)) → c1(F(c(f(z0))), F(z0))
F(f(z0)) → c2(F(d(f(z0))), F(z0))
G(h(z0)) → c7(G(z0))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2, c7

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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