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(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(s(z0), z1)) → f(c(z0, s(z1)))
g(c(z0, s(z1))) → g(c(s(z0), z1))

Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

S tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

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(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

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

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

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

POL(F(x₁)) = 0   
POL(G(x₁)) = [4]x₁   
POL(c(x₁, x₂)) = x₂   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [2] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(s(z0), z1)) → f(c(z0, s(z1)))
g(c(z0, s(z1))) → g(c(s(z0), z1))

Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

S tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))

K tuples:

G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

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.

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))

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

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

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

POL(F(x₁)) = [4]x₁   
POL(G(x₁)) = 0   
POL(c(x₁, x₂)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [4] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(s(z0), z1)) → f(c(z0, s(z1)))
g(c(z0, s(z1))) → g(c(s(z0), z1))

Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

S tuples:none
K tuples:

G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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