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(0) → 0
f(s(x)) → f(id(x))
id(0) → 0
id(s(x)) → s(id(x))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → 0
f(s(z0)) → f(id(z0))
id(0) → 0
id(s(z0)) → s(id(z0))

Tuples:

F(s(z0)) → c1(F(id(z0)), ID(z0))
ID(s(z0)) → c3(ID(z0))

S tuples:

F(s(z0)) → c1(F(id(z0)), ID(z0))
ID(s(z0)) → c3(ID(z0))

K tuples:none
Defined Rule Symbols:

f, id

Defined Pair Symbols:

F, ID

Compound Symbols:

c1, c3

(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.

F(s(z0)) → c1(F(id(z0)), ID(z0))

We considered the (Usable) Rules:

id(0) → 0
id(s(z0)) → s(id(z0))

And the Tuples:

F(s(z0)) → c1(F(id(z0)), ID(z0))
ID(s(z0)) → c3(ID(z0))

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

POL(0) = 0   
POL(F(x₁)) = x₁   
POL(ID(x₁)) = 0   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(id(x₁)) = x₁   
POL(s(x₁)) = [2] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → 0
f(s(z0)) → f(id(z0))
id(0) → 0
id(s(z0)) → s(id(z0))

Tuples:

F(s(z0)) → c1(F(id(z0)), ID(z0))
ID(s(z0)) → c3(ID(z0))

S tuples:

ID(s(z0)) → c3(ID(z0))

K tuples:

F(s(z0)) → c1(F(id(z0)), ID(z0))

Defined Rule Symbols:

f, id

Defined Pair Symbols:

F, ID

Compound Symbols:

c1, c3

(5) 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.

ID(s(z0)) → c3(ID(z0))

We considered the (Usable) Rules:

id(0) → 0
id(s(z0)) → s(id(z0))

And the Tuples:

F(s(z0)) → c1(F(id(z0)), ID(z0))
ID(s(z0)) → c3(ID(z0))

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

POL(0) = 0   
POL(F(x₁)) = x₁²   
POL(ID(x₁)) = [1] + [2]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(id(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → 0
f(s(z0)) → f(id(z0))
id(0) → 0
id(s(z0)) → s(id(z0))

Tuples:

F(s(z0)) → c1(F(id(z0)), ID(z0))
ID(s(z0)) → c3(ID(z0))

S tuples:none
K tuples:

F(s(z0)) → c1(F(id(z0)), ID(z0))
ID(s(z0)) → c3(ID(z0))

Defined Rule Symbols:

f, id

Defined Pair Symbols:

F, ID

Compound Symbols:

c1, c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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