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:

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

I(+(z0, z1)) → c2(+'(i(z0), i(z1)), I(z0), I(z1))
+'(z0, +(z1, z2)) → c7(+'(+(z0, z1), z2), +'(z0, z1))

S tuples:

I(+(z0, z1)) → c2(+'(i(z0), i(z1)), I(z0), I(z1))
+'(z0, +(z1, z2)) → c7(+'(+(z0, z1), z2), +'(z0, z1))

K tuples:none
Defined Rule Symbols:

i, +

Defined Pair Symbols:

I, +'

Compound Symbols:

c2, c7

(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.

I(+(z0, z1)) → c2(+'(i(z0), i(z1)), I(z0), I(z1))

We considered the (Usable) Rules:

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

And the Tuples:

I(+(z0, z1)) → c2(+'(i(z0), i(z1)), I(z0), I(z1))
+'(z0, +(z1, z2)) → c7(+'(+(z0, z1), z2), +'(z0, z1))

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

POL(+(x₁, x₂)) = [3] + [4]x₁ + [3]x₂   
POL(+'(x₁, x₂)) = 0   
POL(0) = [5]   
POL(I(x₁)) = [5] + [2]x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(i(x₁)) = [4] + [5]x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

I(+(z0, z1)) → c2(+'(i(z0), i(z1)), I(z0), I(z1))
+'(z0, +(z1, z2)) → c7(+'(+(z0, z1), z2), +'(z0, z1))

S tuples:

+'(z0, +(z1, z2)) → c7(+'(+(z0, z1), z2), +'(z0, z1))

K tuples:

I(+(z0, z1)) → c2(+'(i(z0), i(z1)), I(z0), I(z1))

Defined Rule Symbols:

i, +

Defined Pair Symbols:

I, +'

Compound Symbols:

c2, c7

(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)) → c7(+'(+(z0, z1), z2), +'(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

I(+(z0, z1)) → c2(+'(i(z0), i(z1)), I(z0), I(z1))
+'(z0, +(z1, z2)) → c7(+'(+(z0, z1), z2), +'(z0, z1))

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

POL(+(x₁, x₂)) = [2] + [2]x₁ + [3]x₂   
POL(+'(x₁, x₂)) = [4]x₂   
POL(0) = [2]   
POL(I(x₁)) = [4] + [2]x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(i(x₁)) = x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

I(+(z0, z1)) → c2(+'(i(z0), i(z1)), I(z0), I(z1))
+'(z0, +(z1, z2)) → c7(+'(+(z0, z1), z2), +'(z0, z1))

S tuples:none
K tuples:

I(+(z0, z1)) → c2(+'(i(z0), i(z1)), I(z0), I(z1))
+'(z0, +(z1, z2)) → c7(+'(+(z0, z1), z2), +'(z0, z1))

Defined Rule Symbols:

i, +

Defined Pair Symbols:

I, +'

Compound Symbols:

c2, c7

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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