### (0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

The TRS R consists of the following rules:

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS is a non-duplicating overlay system, we have rc = irc.

### (2) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

The TRS R consists of the following rules:

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil))
car(.(z0, z1)) → z0
cdr(.(z0, z1)) → z1
null(nil) → true
null(.(z0, z1)) → false
++(nil, z0) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
Tuples:

REV(nil) → c
REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
CAR(.(z0, z1)) → c2
CDR(.(z0, z1)) → c3
NULL(nil) → c4
NULL(.(z0, z1)) → c5
++'(nil, z0) → c6
++'(.(z0, z1), z2) → c7(++'(z1, z2))
S tuples:

REV(nil) → c
REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
CAR(.(z0, z1)) → c2
CDR(.(z0, z1)) → c3
NULL(nil) → c4
NULL(.(z0, z1)) → c5
++'(nil, z0) → c6
++'(.(z0, z1), z2) → c7(++'(z1, z2))
K tuples:none
Defined Rule Symbols:

rev, car, cdr, null, ++

Defined Pair Symbols:

REV, CAR, CDR, NULL, ++'

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7

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

Removed 6 trailing nodes:

REV(nil) → c
NULL(nil) → c4
++'(nil, z0) → c6
CDR(.(z0, z1)) → c3
CAR(.(z0, z1)) → c2
NULL(.(z0, z1)) → c5

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil))
car(.(z0, z1)) → z0
cdr(.(z0, z1)) → z1
null(nil) → true
null(.(z0, z1)) → false
++(nil, z0) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))
S tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))
K tuples:none
Defined Rule Symbols:

rev, car, cdr, null, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

car(.(z0, z1)) → z0
cdr(.(z0, z1)) → z1
null(nil) → true
null(.(z0, z1)) → false

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil))
++(nil, z0) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))
S tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))
K tuples:none
Defined Rule Symbols:

rev, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

### (9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
We considered the (Usable) Rules:none
And the Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(++(x1, x2)) = 0
POL(++'(x1, x2)) = 0
POL(.(x1, x2)) = [1] + x2
POL(REV(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(nil) = 0
POL(rev(x1)) = 0

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil))
++(nil, z0) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))
S tuples:

++'(.(z0, z1), z2) → c7(++'(z1, z2))
K tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
Defined Rule Symbols:

rev, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

### (11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

++'(.(z0, z1), z2) → c7(++'(z1, z2))
We considered the (Usable) Rules:

rev(.(z0, z1)) → ++(rev(z1), .(z0, nil))
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
rev(nil) → nil
++(nil, z0) → z0
And the Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(++(x1, x2)) = x1 + [2]x2
POL(++'(x1, x2)) = [1] + [2]x1
POL(.(x1, x2)) = [2] + x2
POL(REV(x1)) = x12
POL(c1(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(nil) = 0
POL(rev(x1)) = [2]x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil))
++(nil, z0) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))
S tuples:none
K tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))
Defined Rule Symbols:

rev, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

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

The set S is empty