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(x, 0) → s(0)
f(s(x), s(y)) → s(f(x, y))
g(0, x) → g(f(x, x), x)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)

Tuples:

F(z0, 0) → c
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

S tuples:

F(z0, 0) → c
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2

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

Removed 1 trailing nodes:

F(z0, 0) → c

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)

Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

g(0, z0) → g(f(z0, z0), z0)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))

Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(7) CdtRuleRemovalProof (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(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

We considered the (Usable) Rules:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))

And the Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

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

POL(0) = [4]   
POL(F(x₁, x₂)) = [3]   
POL(G(x₁, x₂)) = [4]x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(f(x₁, x₂)) = 0   
POL(s(x₁)) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))

Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

K tuples:

G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

Defined Rule Symbols:

f

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(9) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0)) by

G(0, 0) → c2(G(s(0), 0), F(0, 0))
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))

Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, 0) → c2(G(s(0), 0), F(0, 0))
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))

S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

K tuples:

G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))

Defined Rule Symbols:

f

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

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

Removed 1 trailing nodes:

G(0, 0) → c2(G(s(0), 0), F(0, 0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))

Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))

S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

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

Removed 1 trailing tuple parts

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))

Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, s(z0)) → c2(F(s(z0), s(z0)))

S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(15) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

G(0, s(z0)) → c2(F(s(z0), s(z0)))

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))

Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

(17) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c1

(19) CdtRuleRemovalProof (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), s(z1)) → c1(F(z0, z1))

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

F(s(z0), s(z1)) → c1(F(z0, z1))

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

POL(F(x₁, x₂)) = [2]x₁   
POL(c1(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

S tuples:none
K tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c1

(21) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(22) BOUNDS(O(1), O(1))