The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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))
+(0, s(z0)) → s(z0)
s(+(0, z0)) → s(z0)

Tuples:

+'(z0, s(z1)) → c1(S(+(z0, z1)), +'(z0, z1))
+'(0, s(z0)) → c2(S(z0))
S(+(0, z0)) → c3(S(z0))

S tuples:

+'(z0, s(z1)) → c1(S(+(z0, z1)), +'(z0, z1))
+'(0, s(z0)) → c2(S(z0))
S(+(0, z0)) → c3(S(z0))

K tuples:none
Defined Rule Symbols:

+, s

Defined Pair Symbols:

+', S

Compound Symbols:

c1, c2, c3

(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.

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

We considered the (Usable) Rules:

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

And the Tuples:

+'(z0, s(z1)) → c1(S(+(z0, z1)), +'(z0, z1))
+'(0, s(z0)) → c2(S(z0))
S(+(0, z0)) → c3(S(z0))

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

POL(+(x₁, x₂)) = [3] + [5]x₂   
POL(+'(x₁, x₂)) = [3]x₁ + [4]x₂   
POL(0) = 0   
POL(S(x₁)) = [5]   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(s(x₁)) = [2] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

+'(z0, s(z1)) → c1(S(+(z0, z1)), +'(z0, z1))
+'(0, s(z0)) → c2(S(z0))
S(+(0, z0)) → c3(S(z0))

S tuples:

S(+(0, z0)) → c3(S(z0))

K tuples:

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

Defined Rule Symbols:

+, s

Defined Pair Symbols:

+', S

Compound Symbols:

c1, c2, c3

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

S(+(0, z0)) → c3(S(z0))

We considered the (Usable) Rules:

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

And the Tuples:

+'(z0, s(z1)) → c1(S(+(z0, z1)), +'(z0, z1))
+'(0, s(z0)) → c2(S(z0))
S(+(0, z0)) → c3(S(z0))

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

POL(+(x₁, x₂)) = [1] + [2]x₂² + x₁·x₂   
POL(+'(x₁, x₂)) = [2]x₁ + [2]x₂² + [2]x₁·x₂ + [3]x₁²   
POL(0) = [2]   
POL(S(x₁)) = [1] + [2]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(s(x₁)) = [2] + [2]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

+'(z0, s(z1)) → c1(S(+(z0, z1)), +'(z0, z1))
+'(0, s(z0)) → c2(S(z0))
S(+(0, z0)) → c3(S(z0))

S tuples:none
K tuples:

+'(z0, s(z1)) → c1(S(+(z0, z1)), +'(z0, z1))
+'(0, s(z0)) → c2(S(z0))
S(+(0, z0)) → c3(S(z0))

Defined Rule Symbols:

+, s

Defined Pair Symbols:

+', S

Compound Symbols:

c1, c2, c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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