### (0) Obligation:

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

rev(ls) → r1(ls, empty)
r1(empty, a) → a
r1(cons(x, k), a) → r1(k, cons(x, a))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(z0) → r1(z0, empty)
r1(empty, z0) → z0
r1(cons(z0, z1), z2) → r1(z1, cons(z0, z2))
Tuples:

REV(z0) → c(R1(z0, empty))
R1(empty, z0) → c1
R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
S tuples:

REV(z0) → c(R1(z0, empty))
R1(empty, z0) → c1
R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
K tuples:none
Defined Rule Symbols:

rev, r1

Defined Pair Symbols:

REV, R1

Compound Symbols:

c, c1, c2

### (3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

REV(z0) → c(R1(z0, empty))
Removed 1 trailing nodes:

R1(empty, z0) → c1

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(z0) → r1(z0, empty)
r1(empty, z0) → z0
r1(cons(z0, z1), z2) → r1(z1, cons(z0, z2))
Tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
S tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
K tuples:none
Defined Rule Symbols:

rev, r1

Defined Pair Symbols:

R1

Compound Symbols:

c2

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

rev(z0) → r1(z0, empty)
r1(empty, z0) → z0
r1(cons(z0, z1), z2) → r1(z1, cons(z0, z2))

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
S tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

R1

Compound Symbols:

c2

### (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.

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
We considered the (Usable) Rules:none
And the Tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(R1(x1, x2)) = [4]x1 + [3]x2
POL(c2(x1)) = x1
POL(cons(x1, x2)) = [5] + x1 + x2

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
S tuples:none
K tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))
Defined Rule Symbols:none

Defined Pair Symbols:

R1

Compound Symbols:

c2

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

The set S is empty