### (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 Cpx (relative) TRS 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:

++'(nil, z0) → c
++'(z0, nil) → c1
++'(.(z0, z1), z2) → c2(++'(z1, z2))
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))
S tuples:

++'(nil, z0) → c
++'(z0, nil) → c1
++'(.(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:

c, c1, c2, c3

### (3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing nodes:

++'(nil, z0) → c
++'(z0, nil) → c1

### (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) → 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

### (5) CdtRuleRemovalProof (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))
++'(++(z0, z1), z2) → c3(++'(z0, ++(z1, z2)), ++'(z1, z2))
We considered the (Usable) Rules:none
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

### (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 (BOTH BOUNDS(ID, ID) transformation)

The set S is empty