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:

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

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c, c1, c2

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

Removed 2 trailing nodes:

+'(0, z0) → c
+'(s(z0), 0) → c1

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), 0) → s(z0)

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

(9) 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.

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

We considered the (Usable) Rules:

+(s(z0), 0) → s(z0)
+(0, z0) → z0

And the Tuples:

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

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

POL(+(x₁, x₂)) = x₁ + [4]x₂   
POL(+'(x₁, x₂)) = [4]x₂   
POL(0) = 0   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [4] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), 0) → s(z0)

Tuples:

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

S tuples:none
K tuples:

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

Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

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

The set S is empty

(12) BOUNDS(O(1), O(1))