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

(0) Obligation:

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

A__F(s(0)) → c1(A__F(a__p(s(0))), A__P(s(0)))
MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(p(z0)) → c6(A__P(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))

S tuples:

A__F(s(0)) → c1(A__F(a__p(s(0))), A__P(s(0)))
MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(p(z0)) → c6(A__P(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))

K tuples:none
Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c1, c5, c6, c8, c9

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

Removed 2 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))
MARK(p(z0)) → c6(MARK(z0))

S tuples:

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))
MARK(p(z0)) → c6(MARK(z0))

K tuples:none
Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c5, c8, c9, c1, c6

(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.

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))

We considered the (Usable) Rules:

a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))
a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)

And the Tuples:

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))
MARK(p(z0)) → c6(MARK(z0))

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

POL(0) = 0   
POL(A__F(x₁)) = 0   
POL(MARK(x₁)) = [4]x₁   
POL(a__f(x₁)) = [3] + [2]x₁   
POL(a__p(x₁)) = [2]   
POL(c1(x₁)) = x₁   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(cons(x₁, x₂)) = [4] + x₁   
POL(f(x₁)) = [5] + x₁   
POL(mark(x₁)) = [2] + [2]x₁   
POL(p(x₁)) = [5] + x₁   
POL(s(x₁)) = x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))
MARK(p(z0)) → c6(MARK(z0))

S tuples:

MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))

K tuples:

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c5, c8, c9, c1, c6

(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(s(z0)) → c9(MARK(z0))

We considered the (Usable) Rules:

a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))
a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)

And the Tuples:

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))
MARK(p(z0)) → c6(MARK(z0))

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

POL(0) = 0   
POL(A__F(x₁)) = 0   
POL(MARK(x₁)) = [3]x₁   
POL(a__f(x₁)) = [3] + [2]x₁   
POL(a__p(x₁)) = 0   
POL(c1(x₁)) = x₁   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(cons(x₁, x₂)) = [5] + x₁   
POL(f(x₁)) = x₁   
POL(mark(x₁)) = [2] + [3]x₁   
POL(p(x₁)) = [2] + x₁   
POL(s(x₁)) = [3] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))
MARK(p(z0)) → c6(MARK(z0))

S tuples:

A__F(s(0)) → c1(A__F(a__p(s(0))))

K tuples:

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c5, c8, c9, c1, c6

(9) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0)) by

MARK(f(f(z0))) → c5(A__F(a__f(mark(z0))), MARK(f(z0)))
MARK(f(p(z0))) → c5(A__F(a__p(mark(z0))), MARK(p(z0)))
MARK(f(0)) → c5(A__F(0), MARK(0))
MARK(f(cons(z0, z1))) → c5(A__F(cons(mark(z0), z1)), MARK(cons(z0, z1)))
MARK(f(s(z0))) → c5(A__F(s(mark(z0))), MARK(s(z0)))
MARK(f(x0)) → c5

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))
MARK(p(z0)) → c6(MARK(z0))
MARK(f(f(z0))) → c5(A__F(a__f(mark(z0))), MARK(f(z0)))
MARK(f(p(z0))) → c5(A__F(a__p(mark(z0))), MARK(p(z0)))
MARK(f(0)) → c5(A__F(0), MARK(0))
MARK(f(cons(z0, z1))) → c5(A__F(cons(mark(z0), z1)), MARK(cons(z0, z1)))
MARK(f(s(z0))) → c5(A__F(s(mark(z0))), MARK(s(z0)))
MARK(f(x0)) → c5

S tuples:

A__F(s(0)) → c1(A__F(a__p(s(0))))

K tuples:

MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c8, c9, c1, c6, c5, c5

(11) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing nodes:

MARK(f(x0)) → c5
MARK(f(0)) → c5(A__F(0), MARK(0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))
MARK(p(z0)) → c6(MARK(z0))
MARK(f(f(z0))) → c5(A__F(a__f(mark(z0))), MARK(f(z0)))
MARK(f(p(z0))) → c5(A__F(a__p(mark(z0))), MARK(p(z0)))
MARK(f(cons(z0, z1))) → c5(A__F(cons(mark(z0), z1)), MARK(cons(z0, z1)))
MARK(f(s(z0))) → c5(A__F(s(mark(z0))), MARK(s(z0)))

S tuples:

A__F(s(0)) → c1(A__F(a__p(s(0))))

K tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c8, c9, c1, c6, c5

(13) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1(A__F(a__p(s(0))))
MARK(p(z0)) → c6(MARK(z0))
MARK(f(f(z0))) → c5(A__F(a__f(mark(z0))), MARK(f(z0)))
MARK(f(p(z0))) → c5(A__F(a__p(mark(z0))), MARK(p(z0)))
MARK(f(s(z0))) → c5(A__F(s(mark(z0))), MARK(s(z0)))
MARK(f(cons(z0, z1))) → c5(MARK(cons(z0, z1)))

S tuples:

A__F(s(0)) → c1(A__F(a__p(s(0))))

K tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c8, c9, c1, c6, c5, c5

(15) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace A__F(s(0)) → c1(A__F(a__p(s(0)))) by

A__F(s(0)) → c1(A__F(0))
A__F(s(0)) → c1(A__F(p(s(0))))
A__F(s(0)) → c1

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(f(f(z0))) → c5(A__F(a__f(mark(z0))), MARK(f(z0)))
MARK(f(p(z0))) → c5(A__F(a__p(mark(z0))), MARK(p(z0)))
MARK(f(s(z0))) → c5(A__F(s(mark(z0))), MARK(s(z0)))
MARK(f(cons(z0, z1))) → c5(MARK(cons(z0, z1)))
A__F(s(0)) → c1(A__F(0))
A__F(s(0)) → c1(A__F(p(s(0))))
A__F(s(0)) → c1

S tuples:

A__F(s(0)) → c1(A__F(0))
A__F(s(0)) → c1(A__F(p(s(0))))
A__F(s(0)) → c1

K tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c8, c9, c6, c5, c5, c1, c1

(17) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing tuple parts

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(f(f(z0))) → c5(A__F(a__f(mark(z0))), MARK(f(z0)))
MARK(f(p(z0))) → c5(A__F(a__p(mark(z0))), MARK(p(z0)))
MARK(f(s(z0))) → c5(A__F(s(mark(z0))), MARK(s(z0)))
MARK(f(cons(z0, z1))) → c5(MARK(cons(z0, z1)))
A__F(s(0)) → c1

S tuples:

A__F(s(0)) → c1

K tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c8, c9, c6, c5, c5, c1

(19) 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__F(s(0)) → c1

We considered the (Usable) Rules:

mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))
a__p(s(0)) → 0
a__p(z0) → p(z0)
a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)

And the Tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(f(f(z0))) → c5(A__F(a__f(mark(z0))), MARK(f(z0)))
MARK(f(p(z0))) → c5(A__F(a__p(mark(z0))), MARK(p(z0)))
MARK(f(s(z0))) → c5(A__F(s(mark(z0))), MARK(s(z0)))
MARK(f(cons(z0, z1))) → c5(MARK(cons(z0, z1)))
A__F(s(0)) → c1

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

POL(0) = 0   
POL(A__F(x₁)) = [1]   
POL(MARK(x₁)) = x₁   
POL(a__f(x₁)) = 0   
POL(a__p(x₁)) = 0   
POL(c1) = 0   
POL(c5(x₁)) = x₁   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁   
POL(f(x₁)) = [1] + x₁   
POL(mark(x₁)) = 0   
POL(p(x₁)) = x₁   
POL(s(x₁)) = x₁   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__f(z0) → f(z0)
a__p(s(0)) → 0
a__p(z0) → p(z0)
mark(f(z0)) → a__f(mark(z0))
mark(p(z0)) → a__p(mark(z0))
mark(0) → 0
mark(cons(z0, z1)) → cons(mark(z0), z1)
mark(s(z0)) → s(mark(z0))

Tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(f(f(z0))) → c5(A__F(a__f(mark(z0))), MARK(f(z0)))
MARK(f(p(z0))) → c5(A__F(a__p(mark(z0))), MARK(p(z0)))
MARK(f(s(z0))) → c5(A__F(s(mark(z0))), MARK(s(z0)))
MARK(f(cons(z0, z1))) → c5(MARK(cons(z0, z1)))
A__F(s(0)) → c1

S tuples:none
K tuples:

MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(p(z0)) → c6(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))
A__F(s(0)) → c1

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c8, c9, c6, c5, c5, c1

(21) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(22) BOUNDS(O(1), O(1))