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)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(+(z0, 0)) → f(z0)
+(z0, +(z1, z2)) → +(+(z0, z1), z2)

Tuples:

F(+(z0, 0)) → c(F(z0))
+'(z0, +(z1, z2)) → c1(+'(+(z0, z1), z2), +'(z0, z1))

S tuples:

F(+(z0, 0)) → c(F(z0))
+'(z0, +(z1, z2)) → c1(+'(+(z0, z1), z2), +'(z0, z1))

K tuples:none
Defined Rule Symbols:

f, +

Defined Pair Symbols:

F, +'

Compound Symbols:

c, c1

(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(+(z0, 0)) → c(F(z0))

We considered the (Usable) Rules:

+(z0, +(z1, z2)) → +(+(z0, z1), z2)

And the Tuples:

F(+(z0, 0)) → c(F(z0))
+'(z0, +(z1, z2)) → c1(+'(+(z0, z1), z2), +'(z0, z1))

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

POL(+(x₁, x₂)) = [1] + [2]x₁   
POL(+'(x₁, x₂)) = 0   
POL(0) = [4]   
POL(F(x₁)) = x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(+(z0, 0)) → f(z0)
+(z0, +(z1, z2)) → +(+(z0, z1), z2)

Tuples:

F(+(z0, 0)) → c(F(z0))
+'(z0, +(z1, z2)) → c1(+'(+(z0, z1), z2), +'(z0, z1))

S tuples:

+'(z0, +(z1, z2)) → c1(+'(+(z0, z1), z2), +'(z0, z1))

K tuples:

F(+(z0, 0)) → c(F(z0))

Defined Rule Symbols:

f, +

Defined Pair Symbols:

F, +'

Compound Symbols:

c, c1

(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.

+'(z0, +(z1, z2)) → c1(+'(+(z0, z1), z2), +'(z0, z1))

We considered the (Usable) Rules:

+(z0, +(z1, z2)) → +(+(z0, z1), z2)

And the Tuples:

F(+(z0, 0)) → c(F(z0))
+'(z0, +(z1, z2)) → c1(+'(+(z0, z1), z2), +'(z0, z1))

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

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

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(+(z0, 0)) → f(z0)
+(z0, +(z1, z2)) → +(+(z0, z1), z2)

Tuples:

F(+(z0, 0)) → c(F(z0))
+'(z0, +(z1, z2)) → c1(+'(+(z0, z1), z2), +'(z0, z1))

S tuples:none
K tuples:

F(+(z0, 0)) → c(F(z0))
+'(z0, +(z1, z2)) → c1(+'(+(z0, z1), z2), +'(z0, z1))

Defined Rule Symbols:

f, +

Defined Pair Symbols:

F, +'

Compound Symbols:

c, c1

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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