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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 183 ms)
↳4 CdtProblem
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 47 ms)
↳6 CdtProblem
↳7 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳8 CdtProblem
↳9 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳10 CdtProblem
↳11 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 CdtProblem
↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 3 ms)
↳14 CdtProblem
↳15 SIsEmptyProof (⇔, 0 ms)
↳16 BOUNDS(O(1), O(1))

(0) Obligation:

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

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

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(z0)
f(s(z0)) → f(z0)
g(s(0)) → g(f(s(0)))
Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(G(f(s(0))), F(s(0)))
S tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(G(f(s(0))), F(s(0)))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, 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.

F(f(z0)) → c(F(z0))
We considered the (Usable) Rules:

f(s(z0)) → f(z0)
f(f(z0)) → f(z0)
And the Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(G(f(s(0))), F(s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

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

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(z0)
f(s(z0)) → f(z0)
g(s(0)) → g(f(s(0)))
Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(G(f(s(0))), F(s(0)))
S tuples:

F(s(z0)) → c1(F(z0))
G(s(0)) → c2(G(f(s(0))), F(s(0)))
K tuples:

F(f(z0)) → c(F(z0))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, 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.

G(s(0)) → c2(G(f(s(0))), F(s(0)))
We considered the (Usable) Rules:

f(s(z0)) → f(z0)
f(f(z0)) → f(z0)
And the Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(G(f(s(0))), F(s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [4]   
POL(F(x1)) = [3]   
POL(G(x1)) = [2]x1   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c2(x1, x2)) = x1 + x2   
POL(f(x1)) = 0   
POL(s(x1)) = x1   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(z0)
f(s(z0)) → f(z0)
g(s(0)) → g(f(s(0)))
Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(G(f(s(0))), F(s(0)))
S tuples:

F(s(z0)) → c1(F(z0))
K tuples:

F(f(z0)) → c(F(z0))
G(s(0)) → c2(G(f(s(0))), F(s(0)))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2

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

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

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(z0)
f(s(z0)) → f(z0)
g(s(0)) → g(f(s(0)))
Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(G(f(0)), F(s(0)))
G(s(0)) → c2
S tuples:

F(s(z0)) → c1(F(z0))
K tuples:

F(f(z0)) → c(F(z0))
G(s(0)) → c2(G(f(s(0))), F(s(0)))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2, c2

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

Removed 1 trailing nodes:

G(s(0)) → c2

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(z0)
f(s(z0)) → f(z0)
g(s(0)) → g(f(s(0)))
Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(G(f(0)), F(s(0)))
S tuples:

F(s(z0)) → c1(F(z0))
K tuples:

F(f(z0)) → c(F(z0))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2

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

Removed 1 trailing tuple parts

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(z0)
f(s(z0)) → f(z0)
g(s(0)) → g(f(s(0)))
Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(F(s(0)))
S tuples:

F(s(z0)) → c1(F(z0))
K tuples:

F(f(z0)) → c(F(z0))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2

(13) 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(z0))
We considered the (Usable) Rules:none
And the Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(F(s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [5]   
POL(F(x1)) = [5] + x1   
POL(G(x1)) = [4] + [5]x1   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(f(x1)) = [4]x1   
POL(s(x1)) = [1] + x1   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(z0)
f(s(z0)) → f(z0)
g(s(0)) → g(f(s(0)))
Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
G(s(0)) → c2(F(s(0)))
S tuples:none
K tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2

(15) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(16) BOUNDS(O(1), O(1))