The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

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

f(a) → g(h(a))
h(g(x)) → g(h(f(x)))
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(x)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(a) → c(H(a))
H(g(z0)) → c1(H(f(z0)), F(z0))
K(z0, h(z0), a) → c2(H(z0))
K(f(z0), z1, z0) → c3(F(z0))

S tuples:

F(a) → c(H(a))
H(g(z0)) → c1(H(f(z0)), F(z0))
K(z0, h(z0), a) → c2(H(z0))
K(f(z0), z1, z0) → c3(F(z0))

K tuples:none
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

F, H, K

Compound Symbols:

c, c1, c2, c3

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

K(z0, h(z0), a) → c2(H(z0))
K(f(z0), z1, z0) → c3(F(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(a) → c(H(a))
H(g(z0)) → c1(H(f(z0)), F(z0))

S tuples:

F(a) → c(H(a))
H(g(z0)) → c1(H(f(z0)), F(z0))

K tuples:none
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

F, H

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

F(a) → c(H(a))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

H(g(z0)) → c1(H(f(z0)), F(z0))

S tuples:

H(g(z0)) → c1(H(f(z0)), F(z0))

K tuples:none
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

H

Compound Symbols:

c1

(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

H(g(z0)) → c1(H(f(z0)))

S tuples:

H(g(z0)) → c1(H(f(z0)))

K tuples:none
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

H

Compound Symbols:

c1

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

H(g(z0)) → c1(H(f(z0)))

We considered the (Usable) Rules:

f(a) → g(h(a))

And the Tuples:

H(g(z0)) → c1(H(f(z0)))

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

POL(H(x₁)) = x₁ + x₁²   
POL(a) = [1]   
POL(c1(x₁)) = x₁   
POL(f(x₁)) = x₁   
POL(g(x₁)) = [1] + x₁   
POL(h(x₁)) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

H(g(z0)) → c1(H(f(z0)))

S tuples:none
K tuples:

H(g(z0)) → c1(H(f(z0)))

Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

H

Compound Symbols:

c1

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(12) BOUNDS(O(1), O(1))