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:

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(x))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

.(1, z0) → z0
.(z0, 1) → z0
.(i(z0), z0) → 1
.(z0, i(z0)) → 1
.(i(z0), .(z0, z1)) → z1
.(z0, .(i(z0), z1)) → z1
.(.(z0, z1), z2) → .(z0, .(z1, z2))
i(1) → 1
i(i(z0)) → z0
i(.(z0, z1)) → .(i(z1), i(z0))

Tuples:

.'(.(z0, z1), z2) → c6(.'(z0, .(z1, z2)), .'(z1, z2))
I(.(z0, z1)) → c9(.'(i(z1), i(z0)), I(z1), I(z0))

S tuples:

.'(.(z0, z1), z2) → c6(.'(z0, .(z1, z2)), .'(z1, z2))
I(.(z0, z1)) → c9(.'(i(z1), i(z0)), I(z1), I(z0))

K tuples:none
Defined Rule Symbols:

., i

Defined Pair Symbols:

.', I

Compound Symbols:

c6, c9

(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)) → c9(.'(i(z1), i(z0)), I(z1), I(z0))

We considered the (Usable) Rules:

i(i(z0)) → z0
i(.(z0, z1)) → .(i(z1), i(z0))
.(1, z0) → z0
.(z0, 1) → z0
.(i(z0), z0) → 1
.(z0, i(z0)) → 1
.(i(z0), .(z0, z1)) → z1
.(z0, .(i(z0), z1)) → z1
.(.(z0, z1), z2) → .(z0, .(z1, z2))

And the Tuples:

.'(.(z0, z1), z2) → c6(.'(z0, .(z1, z2)), .'(z1, z2))
I(.(z0, z1)) → c9(.'(i(z1), i(z0)), I(z1), I(z0))

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

POL(.(x₁, x₂)) = [3] + [4]x₁ + [4]x₂   
POL(.'(x₁, x₂)) = 0   
POL(1) = [2]   
POL(I(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c9(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(i(x₁)) = [1] + [5]x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

.(1, z0) → z0
.(z0, 1) → z0
.(i(z0), z0) → 1
.(z0, i(z0)) → 1
.(i(z0), .(z0, z1)) → z1
.(z0, .(i(z0), z1)) → z1
.(.(z0, z1), z2) → .(z0, .(z1, z2))
i(1) → 1
i(i(z0)) → z0
i(.(z0, z1)) → .(i(z1), i(z0))

Tuples:

.'(.(z0, z1), z2) → c6(.'(z0, .(z1, z2)), .'(z1, z2))
I(.(z0, z1)) → c9(.'(i(z1), i(z0)), I(z1), I(z0))

S tuples:

.'(.(z0, z1), z2) → c6(.'(z0, .(z1, z2)), .'(z1, z2))

K tuples:

I(.(z0, z1)) → c9(.'(i(z1), i(z0)), I(z1), I(z0))

Defined Rule Symbols:

., i

Defined Pair Symbols:

.', I

Compound Symbols:

c6, c9

(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.

.'(.(z0, z1), z2) → c6(.'(z0, .(z1, z2)), .'(z1, z2))

We considered the (Usable) Rules:

i(i(z0)) → z0
i(.(z0, z1)) → .(i(z1), i(z0))
.(1, z0) → z0
.(z0, 1) → z0
.(i(z0), z0) → 1
.(z0, i(z0)) → 1
.(i(z0), .(z0, z1)) → z1
.(z0, .(i(z0), z1)) → z1
.(.(z0, z1), z2) → .(z0, .(z1, z2))

And the Tuples:

.'(.(z0, z1), z2) → c6(.'(z0, .(z1, z2)), .'(z1, z2))
I(.(z0, z1)) → c9(.'(i(z1), i(z0)), I(z1), I(z0))

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

POL(.(x₁, x₂)) = [2] + [2]x₁ + [2]x₂ + x₁·x₂   
POL(.'(x₁, x₂)) = [2]x₁²   
POL(1) = 0   
POL(I(x₁)) = [1] + x₁²   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c9(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(i(x₁)) = x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

.(1, z0) → z0
.(z0, 1) → z0
.(i(z0), z0) → 1
.(z0, i(z0)) → 1
.(i(z0), .(z0, z1)) → z1
.(z0, .(i(z0), z1)) → z1
.(.(z0, z1), z2) → .(z0, .(z1, z2))
i(1) → 1
i(i(z0)) → z0
i(.(z0, z1)) → .(i(z1), i(z0))

Tuples:

.'(.(z0, z1), z2) → c6(.'(z0, .(z1, z2)), .'(z1, z2))
I(.(z0, z1)) → c9(.'(i(z1), i(z0)), I(z1), I(z0))

S tuples:none
K tuples:

I(.(z0, z1)) → c9(.'(i(z1), i(z0)), I(z1), I(z0))
.'(.(z0, z1), z2) → c6(.'(z0, .(z1, z2)), .'(z1, z2))

Defined Rule Symbols:

., i

Defined Pair Symbols:

.', I

Compound Symbols:

c6, c9

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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