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:

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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(-(z0, z1), z2) → -(+(z0, z2), z1)
-(+(z0, z1), z1) → z0

Tuples:

+'(-(z0, z1), z2) → c(-'(+(z0, z2), z1), +'(z0, z2))

S tuples:

+'(-(z0, z1), z2) → c(-'(+(z0, z2), z1), +'(z0, z2))

K tuples:none
Defined Rule Symbols:

+, -

Defined Pair Symbols:

+'

Compound Symbols:

c

(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, z1), z2) → c(-'(+(z0, z2), z1), +'(z0, z2))

We considered the (Usable) Rules:

+(-(z0, z1), z2) → -(+(z0, z2), z1)
-(+(z0, z1), z1) → z0

And the Tuples:

+'(-(z0, z1), z2) → c(-'(+(z0, z2), z1), +'(z0, z2))

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

POL(+(x₁, x₂)) = [1] + [4]x₁   
POL(+'(x₁, x₂)) = [4]x₁   
POL(-(x₁, x₂)) = [4] + [4]x₁ + [3]x₂   
POL(-'(x₁, x₂)) = [2] + x₁ + x₂   
POL(c(x₁, x₂)) = x₁ + x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(-(z0, z1), z2) → -(+(z0, z2), z1)
-(+(z0, z1), z1) → z0

Tuples:

+'(-(z0, z1), z2) → c(-'(+(z0, z2), z1), +'(z0, z2))

S tuples:none
K tuples:

+'(-(z0, z1), z2) → c(-'(+(z0, z2), z1), +'(z0, z2))

Defined Rule Symbols:

+, -

Defined Pair Symbols:

+'

Compound Symbols:

c

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(6) BOUNDS(O(1), O(1))