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 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳8 CdtProblem
↳9 CdtKnowledgeProof (⇔, 0 ms)
↳10 CdtProblem
↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 14 ms)
↳12 CdtProblem
↳13 SIsEmptyProof (⇔, 0 ms)
↳14 BOUNDS(O(1), O(1))

(0) Obligation:

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

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
f(0, s(0), z0) → f(z0, +(z0, z0), z0)
g(z0, z1) → z0
g(z0, z1) → z1
Tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), z0) → c2(F(z0, +(z0, z0), z0), +'(z0, z0))
S tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), z0) → c2(F(z0, +(z0, z0), z0), +'(z0, z0))
K tuples:none
Defined Rule Symbols:

+, f, g

Defined Pair Symbols:

+', F

Compound Symbols:

c1, c2

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

Use narrowing to replace F(0, s(0), z0) → c2(F(z0, +(z0, z0), z0), +'(z0, z0)) by

F(0, s(0), 0) → c2(F(0, 0, 0), +'(0, 0))
F(0, s(0), s(z1)) → c2(F(s(z1), s(+(s(z1), z1)), s(z1)), +'(s(z1), s(z1)))
F(0, s(0), x0) → c2

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
f(0, s(0), z0) → f(z0, +(z0, z0), z0)
g(z0, z1) → z0
g(z0, z1) → z1
Tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), 0) → c2(F(0, 0, 0), +'(0, 0))
F(0, s(0), s(z1)) → c2(F(s(z1), s(+(s(z1), z1)), s(z1)), +'(s(z1), s(z1)))
F(0, s(0), x0) → c2
S tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), 0) → c2(F(0, 0, 0), +'(0, 0))
F(0, s(0), s(z1)) → c2(F(s(z1), s(+(s(z1), z1)), s(z1)), +'(s(z1), s(z1)))
F(0, s(0), x0) → c2
K tuples:none
Defined Rule Symbols:

+, f, g

Defined Pair Symbols:

+', F

Compound Symbols:

c1, c2, c2

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

Removed 2 trailing nodes:

F(0, s(0), x0) → c2
F(0, s(0), 0) → c2(F(0, 0, 0), +'(0, 0))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
f(0, s(0), z0) → f(z0, +(z0, z0), z0)
g(z0, z1) → z0
g(z0, z1) → z1
Tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), s(z1)) → c2(F(s(z1), s(+(s(z1), z1)), s(z1)), +'(s(z1), s(z1)))
S tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), s(z1)) → c2(F(s(z1), s(+(s(z1), z1)), s(z1)), +'(s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:

+, f, g

Defined Pair Symbols:

+', F

Compound Symbols:

c1, c2

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

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
f(0, s(0), z0) → f(z0, +(z0, z0), z0)
g(z0, z1) → z0
g(z0, z1) → z1
Tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), s(z1)) → c2(+'(s(z1), s(z1)))
S tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), s(z1)) → c2(+'(s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:

+, f, g

Defined Pair Symbols:

+', F

Compound Symbols:

c1, c2

(9) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

F(0, s(0), s(z1)) → c2(+'(s(z1), s(z1)))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
f(0, s(0), z0) → f(z0, +(z0, z0), z0)
g(z0, z1) → z0
g(z0, z1) → z1
Tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), s(z1)) → c2(+'(s(z1), s(z1)))
S tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
K tuples:

F(0, s(0), s(z1)) → c2(+'(s(z1), s(z1)))
Defined Rule Symbols:

+, f, g

Defined Pair Symbols:

+', F

Compound Symbols:

c1, c2

(11) 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.

+'(z0, s(z1)) → c1(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), s(z1)) → c2(+'(s(z1), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+'(x1, x2)) = [3] + [2]x1 + x2   
POL(0) = [3]   
POL(F(x1, x2, x3)) = [4] + [4]x1 + [4]x3   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(s(x1)) = [4] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
f(0, s(0), z0) → f(z0, +(z0, z0), z0)
g(z0, z1) → z0
g(z0, z1) → z1
Tuples:

+'(z0, s(z1)) → c1(+'(z0, z1))
F(0, s(0), s(z1)) → c2(+'(s(z1), s(z1)))
S tuples:none
K tuples:

F(0, s(0), s(z1)) → c2(+'(s(z1), s(z1)))
+'(z0, s(z1)) → c1(+'(z0, z1))
Defined Rule Symbols:

+, f, g

Defined Pair Symbols:

+', F

Compound Symbols:

c1, c2

(13) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(14) BOUNDS(O(1), O(1))