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

(0) Obligation:

Runtime Complexity TRS:
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

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

Converted CpxTRS to CDT

(2) 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

(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.

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))

We considered the (Usable) Rules:

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

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)) = [4] + x₂   
POL(REV(x₁)) = [2]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(nil) = [1]   
POL(rev(x₁)) = 0   

(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(.(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, car, cdr, null, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(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(nil) → nil
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil))
++(nil, z0) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))

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

(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:none
K tuples:

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

Defined Rule Symbols:

rev, car, cdr, null, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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