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(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 Cpx (relative) TRS 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) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

g(s(0)) → g(f(s(0)))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → f(z0)
f(f(z0)) → f(z0)

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

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2

(5) 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(f(z0)) → c(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(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   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → f(z0)
f(f(z0)) → f(z0)

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

Defined Pair Symbols:

F, G

Compound Symbols:

c, 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(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   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → f(z0)
f(f(z0)) → f(z0)

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

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2

(9) 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)))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → f(z0)
f(f(z0)) → f(z0)

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))
G(s(0)) → c2(G(f(s(0))), F(s(0)))

Defined Rule Symbols:

f

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(s(z0)) → f(z0)
f(f(z0)) → f(z0)

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))
G(s(0)) → c2(G(f(s(0))), F(s(0)))

Defined Rule Symbols:

f

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2

(13) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

G(s(0)) → c2(F(s(0)))

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → f(z0)
f(f(z0)) → f(z0)

Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))

S tuples:

F(s(z0)) → c1(F(z0))

K tuples:

F(f(z0)) → c(F(z0))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

(15) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(s(z0)) → f(z0)
f(f(z0)) → f(z0)

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))

S tuples:

F(s(z0)) → c1(F(z0))

K tuples:

F(f(z0)) → c(F(z0))

Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c, c1

(17) 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)) → c1(F(z0))

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

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))

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

POL(F(x₁)) = x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(f(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))

S tuples:none
K tuples:

F(f(z0)) → c(F(z0))
F(s(z0)) → c1(F(z0))

Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c, c1

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

The set S is empty

(20) BOUNDS(O(1), O(1))