(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 CpxTRS 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:

+'(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

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

Removed 1 trailing tuple parts

(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)))
S tuples:

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

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

(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.

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)))
We considered the (Usable) Rules:

+(0, z0) → z0
+(s(z0), 0) → s(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   

(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:none
K tuples:

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

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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