### (0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(a) → a
rev(b) → b
rev(++(z0, z1)) → ++(rev(z1), rev(z0))
rev(++(z0, z0)) → rev(z0)
Tuples:

REV(a) → c
REV(b) → c1
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
S tuples:

REV(a) → c
REV(b) → c1
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
K tuples:none
Defined Rule Symbols:

rev

Defined Pair Symbols:

REV

Compound Symbols:

c, c1, c2, c3

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

Removed 2 trailing nodes:

REV(a) → c
REV(b) → c1

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(a) → a
rev(b) → b
rev(++(z0, z1)) → ++(rev(z1), rev(z0))
rev(++(z0, z0)) → rev(z0)
Tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
S tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
K tuples:none
Defined Rule Symbols:

rev

Defined Pair Symbols:

REV

Compound Symbols:

c2, c3

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

rev(a) → a
rev(b) → b
rev(++(z0, z1)) → ++(rev(z1), rev(z0))
rev(++(z0, z0)) → rev(z0)

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
S tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

REV

Compound Symbols:

c2, c3

### (7) 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.

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
We considered the (Usable) Rules:none
And the Tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(++(x1, x2)) = [3] + x1 + x2
POL(REV(x1)) = [2] + [2]x1
POL(c2(x1, x2)) = x1 + x2
POL(c3(x1)) = x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
S tuples:none
K tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

REV

Compound Symbols:

c2, c3

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

The set S is empty