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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 268 ms)
↳4 CdtProblem
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))), 655 ms)
↳6 CdtProblem
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))), 667 ms)
↳8 CdtProblem
↳9 SIsEmptyProof (⇔, 0 ms)
↳10 BOUNDS(O(1), O(1))

(0) Obligation:

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(.(z0, z1), z2) → c5(++'(z1, z2))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
S tuples:

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(.(z0, z1), z2) → c5(++'(z1, z2))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
K tuples:none
Defined Rule Symbols:

rev, ++, make

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c2, c5, c6

(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)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
We considered the (Usable) Rules:

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

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(.(z0, z1), z2) → c5(++'(z1, z2))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(++(x1, x2)) = [4] + [2]x1 + [4]x2   
POL(++'(x1, x2)) = 0   
POL(.(x1, x2)) = [5]   
POL(REV(x1)) = [4]x1   
POL(c2(x1, x2, x3)) = x1 + x2 + x3   
POL(c5(x1)) = x1   
POL(c6(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(rev(x1)) = [4] + [4]x1   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(.(z0, z1), z2) → c5(++'(z1, z2))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
S tuples:

++'(.(z0, z1), z2) → c5(++'(z1, z2))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
K tuples:

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
Defined Rule Symbols:

rev, ++, make

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c2, c5, c6

(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)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
We considered the (Usable) Rules:

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

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(.(z0, z1), z2) → c5(++'(z1, z2))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(++(x1, x2)) = [1] + x1 + x2   
POL(++'(x1, x2)) = x2   
POL(.(x1, x2)) = 0   
POL(REV(x1)) = x12   
POL(c2(x1, x2, x3)) = x1 + x2 + x3   
POL(c5(x1)) = x1   
POL(c6(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(rev(x1)) = [2]x1   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(.(z0, z1), z2) → c5(++'(z1, z2))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
S tuples:

++'(.(z0, z1), z2) → c5(++'(z1, z2))
K tuples:

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
Defined Rule Symbols:

rev, ++, make

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c2, c5, c6

(7) 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) → c5(++'(z1, z2))
We considered the (Usable) Rules:

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

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(.(z0, z1), z2) → c5(++'(z1, z2))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(++(x1, x2)) = [2] + [2]x1 + [2]x2 + x1·x2   
POL(++'(x1, x2)) = [2]x1 + x22 + [2]x1·x2   
POL(.(x1, x2)) = [1] + x2   
POL(REV(x1)) = [1] + [2]x1 + [2]x12   
POL(c2(x1, x2, x3)) = x1 + x2 + x3   
POL(c5(x1)) = x1   
POL(c6(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(rev(x1)) = x1   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(.(z0, z1), z2) → c5(++'(z1, z2))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
S tuples:none
K tuples:

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
++'(.(z0, z1), z2) → c5(++'(z1, z2))
Defined Rule Symbols:

rev, ++, make

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c2, c5, c6

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))