The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^3)).

(0) Obligation:

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

a__eq(0, 0) → true
a__eq(s(X), s(Y)) → a__eq(X, Y)
a__eq(X, Y) → false
a__inf(X) → cons(X, inf(s(X)))
a__take(0, X) → nil
a__take(s(X), cons(Y, L)) → cons(Y, take(X, L))
a__length(nil) → 0
a__length(cons(X, L)) → s(length(L))
mark(eq(X1, X2)) → a__eq(X1, X2)
mark(inf(X)) → a__inf(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(length(X)) → a__length(mark(X))
mark(0) → 0
mark(true) → true
mark(s(X)) → s(X)
mark(false) → false
mark(cons(X1, X2)) → cons(X1, X2)
mark(nil) → nil
a__eq(X1, X2) → eq(X1, X2)
a__inf(X) → inf(X)
a__take(X1, X2) → take(X1, X2)
a__length(X) → length(X)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true
a__eq(s(z0), s(z1)) → a__eq(z0, z1)
a__eq(z0, z1) → false
a__eq(z0, z1) → eq(z0, z1)
a__inf(z0) → cons(z0, inf(s(z0)))
a__inf(z0) → inf(z0)
a__take(0, z0) → nil
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2))
a__take(z0, z1) → take(z0, z1)
a__length(nil) → 0
a__length(cons(z0, z1)) → s(length(z1))
a__length(z0) → length(z0)
mark(eq(z0, z1)) → a__eq(z0, z1)
mark(inf(z0)) → a__inf(mark(z0))
mark(take(z0, z1)) → a__take(mark(z0), mark(z1))
mark(length(z0)) → a__length(mark(z0))
mark(0) → 0
mark(true) → true
mark(s(z0)) → s(z0)
mark(false) → false
mark(cons(z0, z1)) → cons(z0, z1)
mark(nil) → nil

Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(A__INF(mark(z0)), MARK(z0))
MARK(take(z0, z1)) → c14(A__TAKE(mark(z0), mark(z1)), MARK(z0), MARK(z1))
MARK(length(z0)) → c15(A__LENGTH(mark(z0)), MARK(z0))

S tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(A__INF(mark(z0)), MARK(z0))
MARK(take(z0, z1)) → c14(A__TAKE(mark(z0), mark(z1)), MARK(z0), MARK(z1))
MARK(length(z0)) → c15(A__LENGTH(mark(z0)), MARK(z0))

K tuples:none
Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

(3) CdtGraphRemoveTrailingProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true
a__eq(s(z0), s(z1)) → a__eq(z0, z1)
a__eq(z0, z1) → false
a__eq(z0, z1) → eq(z0, z1)
a__inf(z0) → cons(z0, inf(s(z0)))
a__inf(z0) → inf(z0)
a__take(0, z0) → nil
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2))
a__take(z0, z1) → take(z0, z1)
a__length(nil) → 0
a__length(cons(z0, z1)) → s(length(z1))
a__length(z0) → length(z0)
mark(eq(z0, z1)) → a__eq(z0, z1)
mark(inf(z0)) → a__inf(mark(z0))
mark(take(z0, z1)) → a__take(mark(z0), mark(z1))
mark(length(z0)) → a__length(mark(z0))
mark(0) → 0
mark(true) → true
mark(s(z0)) → s(z0)
mark(false) → false
mark(cons(z0, z1)) → cons(z0, z1)
mark(nil) → nil

Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

S tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

K tuples:none
Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

(5) 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.

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))

We considered the (Usable) Rules:none
And the Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(A__EQ(x₁, x₂)) = x₁ + x₂   
POL(MARK(x₁)) = [2]x₁   
POL(c1(x₁)) = x₁   
POL(c12(x₁)) = x₁   
POL(c13(x₁)) = x₁   
POL(c14(x₁, x₂)) = x₁ + x₂   
POL(c15(x₁)) = x₁   
POL(eq(x₁, x₂)) = x₁ + x₂   
POL(inf(x₁)) = [1] + x₁   
POL(length(x₁)) = x₁   
POL(s(x₁)) = [4] + x₁   
POL(take(x₁, x₂)) = x₁ + x₂   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true
a__eq(s(z0), s(z1)) → a__eq(z0, z1)
a__eq(z0, z1) → false
a__eq(z0, z1) → eq(z0, z1)
a__inf(z0) → cons(z0, inf(s(z0)))
a__inf(z0) → inf(z0)
a__take(0, z0) → nil
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2))
a__take(z0, z1) → take(z0, z1)
a__length(nil) → 0
a__length(cons(z0, z1)) → s(length(z1))
a__length(z0) → length(z0)
mark(eq(z0, z1)) → a__eq(z0, z1)
mark(inf(z0)) → a__inf(mark(z0))
mark(take(z0, z1)) → a__take(mark(z0), mark(z1))
mark(length(z0)) → a__length(mark(z0))
mark(0) → 0
mark(true) → true
mark(s(z0)) → s(z0)
mark(false) → false
mark(cons(z0, z1)) → cons(z0, z1)
mark(nil) → nil

Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

S tuples:

MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

K tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))

Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

(7) 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.

MARK(inf(z0)) → c13(MARK(z0))

We considered the (Usable) Rules:none
And the Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(A__EQ(x₁, x₂)) = [5] + x₁   
POL(MARK(x₁)) = [5] + [2]x₁   
POL(c1(x₁)) = x₁   
POL(c12(x₁)) = x₁   
POL(c13(x₁)) = x₁   
POL(c14(x₁, x₂)) = x₁ + x₂   
POL(c15(x₁)) = x₁   
POL(eq(x₁, x₂)) = x₁   
POL(inf(x₁)) = [4] + x₁   
POL(length(x₁)) = x₁   
POL(s(x₁)) = x₁   
POL(take(x₁, x₂)) = [4] + x₁ + x₂   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true
a__eq(s(z0), s(z1)) → a__eq(z0, z1)
a__eq(z0, z1) → false
a__eq(z0, z1) → eq(z0, z1)
a__inf(z0) → cons(z0, inf(s(z0)))
a__inf(z0) → inf(z0)
a__take(0, z0) → nil
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2))
a__take(z0, z1) → take(z0, z1)
a__length(nil) → 0
a__length(cons(z0, z1)) → s(length(z1))
a__length(z0) → length(z0)
mark(eq(z0, z1)) → a__eq(z0, z1)
mark(inf(z0)) → a__inf(mark(z0))
mark(take(z0, z1)) → a__take(mark(z0), mark(z1))
mark(length(z0)) → a__length(mark(z0))
mark(0) → 0
mark(true) → true
mark(s(z0)) → s(z0)
mark(false) → false
mark(cons(z0, z1)) → cons(z0, z1)
mark(nil) → nil

Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

S tuples:

MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

K tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))

Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

(9) 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.

MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))

We considered the (Usable) Rules:none
And the Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(A__EQ(x₁, x₂)) = 0   
POL(MARK(x₁)) = [1] + [2]x₁ + [3]x₁²   
POL(c1(x₁)) = x₁   
POL(c12(x₁)) = x₁   
POL(c13(x₁)) = x₁   
POL(c14(x₁, x₂)) = x₁ + x₂   
POL(c15(x₁)) = x₁   
POL(eq(x₁, x₂)) = x₁ + x₂   
POL(inf(x₁)) = x₁   
POL(length(x₁)) = [3] + x₁   
POL(s(x₁)) = 0   
POL(take(x₁, x₂)) = [3] + x₁ + x₂   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true
a__eq(s(z0), s(z1)) → a__eq(z0, z1)
a__eq(z0, z1) → false
a__eq(z0, z1) → eq(z0, z1)
a__inf(z0) → cons(z0, inf(s(z0)))
a__inf(z0) → inf(z0)
a__take(0, z0) → nil
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2))
a__take(z0, z1) → take(z0, z1)
a__length(nil) → 0
a__length(cons(z0, z1)) → s(length(z1))
a__length(z0) → length(z0)
mark(eq(z0, z1)) → a__eq(z0, z1)
mark(inf(z0)) → a__inf(mark(z0))
mark(take(z0, z1)) → a__take(mark(z0), mark(z1))
mark(length(z0)) → a__length(mark(z0))
mark(0) → 0
mark(true) → true
mark(s(z0)) → s(z0)
mark(false) → false
mark(cons(z0, z1)) → cons(z0, z1)
mark(nil) → nil

Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

S tuples:

MARK(length(z0)) → c15(MARK(z0))

K tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))

Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^3))) transformation)

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

MARK(length(z0)) → c15(MARK(z0))

We considered the (Usable) Rules:none
And the Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(A__EQ(x₁, x₂)) = x₁ + x₂   
POL(MARK(x₁)) = x₁²   
POL(c1(x₁)) = x₁   
POL(c12(x₁)) = x₁   
POL(c13(x₁)) = x₁   
POL(c14(x₁, x₂)) = x₁ + x₂   
POL(c15(x₁)) = x₁   
POL(eq(x₁, x₂)) = [1] + x₁ + x₂   
POL(inf(x₁)) = x₁   
POL(length(x₁)) = [1] + x₁   
POL(s(x₁)) = x₁   
POL(take(x₁, x₂)) = [1] + x₁ + x₂   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true
a__eq(s(z0), s(z1)) → a__eq(z0, z1)
a__eq(z0, z1) → false
a__eq(z0, z1) → eq(z0, z1)
a__inf(z0) → cons(z0, inf(s(z0)))
a__inf(z0) → inf(z0)
a__take(0, z0) → nil
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2))
a__take(z0, z1) → take(z0, z1)
a__length(nil) → 0
a__length(cons(z0, z1)) → s(length(z1))
a__length(z0) → length(z0)
mark(eq(z0, z1)) → a__eq(z0, z1)
mark(inf(z0)) → a__inf(mark(z0))
mark(take(z0, z1)) → a__take(mark(z0), mark(z1))
mark(length(z0)) → a__length(mark(z0))
mark(0) → 0
mark(true) → true
mark(s(z0)) → s(z0)
mark(false) → false
mark(cons(z0, z1)) → cons(z0, z1)
mark(nil) → nil

Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

S tuples:none
K tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1))
MARK(inf(z0)) → c13(MARK(z0))
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1))
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
MARK(length(z0)) → c15(MARK(z0))

Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

(13) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(14) BOUNDS(O(1), O(1))