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:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

++(nil, z0) → z0
++(z0, nil) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
++(++(z0, z1), z2) → ++(z0, ++(z1, z2))

Tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2))
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))

S tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2))
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))

K tuples:none
Defined Rule Symbols:

++

Defined Pair Symbols:

++'

Compound Symbols:

c2, c3

(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) → c2(++'(z1, z2))

We considered the (Usable) Rules:

++(z0, nil) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
++(++(z0, z1), z2) → ++(z0, ++(z1, z2))
++(nil, z0) → z0

And the Tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2))
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))

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

POL(++(x₁, x₂)) = [4] + [2]x₁ + [4]x₂   
POL(++'(x₁, x₂)) = [4] + [2]x₁   
POL(.(x₁, x₂)) = [4] + x₂   
POL(c2(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(nil) = 0   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

++(nil, z0) → z0
++(z0, nil) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
++(++(z0, z1), z2) → ++(z0, ++(z1, z2))

Tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2))
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))

S tuples:

++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))

K tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2))

Defined Rule Symbols:

++

Defined Pair Symbols:

++'

Compound Symbols:

c2, c3

(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) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))

We considered the (Usable) Rules:

++(z0, nil) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
++(++(z0, z1), z2) → ++(z0, ++(z1, z2))
++(nil, z0) → z0

And the Tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2))
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))

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

POL(++(x₁, x₂)) = [4] + x₁ + [2]x₂   
POL(++'(x₁, x₂)) = [5] + [4]x₁   
POL(.(x₁, x₂)) = [3] + x₂   
POL(c2(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(nil) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

++(nil, z0) → z0
++(z0, nil) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
++(++(z0, z1), z2) → ++(z0, ++(z1, z2))

Tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2))
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))

S tuples:none
K tuples:

++'(.(z0, z1), z2) → c2(++'(z1, z2))
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))

Defined Rule Symbols:

++

Defined Pair Symbols:

++'

Compound Symbols:

c2, c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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