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

(0) Obligation:

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


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) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 3.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3]
transitions:
00() → 0
cons0(0, 0) → 0
f0(0) → 0
s0(0) → 0
p0(0) → 0
a__f0(0) → 1
a__p0(0) → 2
mark0(0) → 3
01() → 4
01() → 7
s1(7) → 6
f1(6) → 5
cons1(4, 5) → 1
s1(7) → 9
a__p1(9) → 8
a__f1(8) → 1
01() → 2
mark1(0) → 10
a__f1(10) → 3
mark1(0) → 11
a__p1(11) → 3
01() → 3
mark1(0) → 12
cons1(12, 0) → 3
mark1(0) → 13
s1(13) → 3
f1(0) → 1
p1(0) → 2
02() → 8
a__f1(10) → 10
a__f1(10) → 11
a__f1(10) → 12
a__f1(10) → 13
a__p1(11) → 10
a__p1(11) → 11
a__p1(11) → 12
a__p1(11) → 13
01() → 10
01() → 11
01() → 12
01() → 13
cons1(12, 0) → 10
cons1(12, 0) → 11
cons1(12, 0) → 12
cons1(12, 0) → 13
s1(13) → 10
s1(13) → 11
s1(13) → 12
s1(13) → 13
f2(8) → 1
f2(10) → 3
p2(11) → 3
p2(9) → 8
02() → 14
02() → 17
s2(17) → 16
f2(16) → 15
cons2(14, 15) → 3
cons2(14, 15) → 10
cons2(14, 15) → 11
cons2(14, 15) → 12
cons2(14, 15) → 13
s2(17) → 19
a__p2(19) → 18
a__f2(18) → 3
a__f2(18) → 10
a__f2(18) → 11
a__f2(18) → 12
a__f2(18) → 13
02() → 3
02() → 10
02() → 11
02() → 12
02() → 13
f2(10) → 10
f2(10) → 11
f2(10) → 12
f2(10) → 13
p2(11) → 10
p2(11) → 11
p2(11) → 12
p2(11) → 13
cons2(14, 15) → 1
03() → 18
f3(18) → 3
f3(18) → 10
f3(18) → 11
f3(18) → 12
f3(18) → 13
p3(19) → 18
03() → 20
03() → 23
s3(23) → 22
f3(22) → 21
cons3(20, 21) → 3
cons3(20, 21) → 10
cons3(20, 21) → 11
cons3(20, 21) → 12
cons3(20, 21) → 13

(2) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(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:

A__F(0) → c
A__F(s(0)) → c1(A__F(a__p(s(0))), A__P(s(0)))
A__F(z0) → c2
A__P(s(0)) → c3
A__P(z0) → c4
MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(p(z0)) → c6(A__P(mark(z0)), MARK(z0))
MARK(0) → c7
MARK(cons(z0, z1)) → c8(MARK(z0))
MARK(s(z0)) → c9(MARK(z0))

S tuples:

A__F(0) → c
A__F(s(0)) → c1(A__F(a__p(s(0))), A__P(s(0)))
A__F(z0) → c2
A__P(s(0)) → c3
A__P(z0) → c4
MARK(f(z0)) → c5(A__F(mark(z0)), MARK(z0))
MARK(p(z0)) → c6(A__P(mark(z0)), MARK(z0))
MARK(0) → c7
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, A__P, MARK

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9

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

Removed 5 trailing nodes:

MARK(0) → c7
A__F(0) → c
A__P(s(0)) → c3
A__P(z0) → c4
A__F(z0) → c2

(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:

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

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

Removed 2 trailing tuple parts

(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:

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

(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.

MARK(p(z0)) → c6(MARK(z0))

We considered the (Usable) Rules:none
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)) = [2]   
POL(MARK(x₁)) = [2]x₁   
POL(a__f(x₁)) = [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)) = x₁   
POL(f(x₁)) = [1] + x₁   
POL(mark(x₁)) = [2]x₁   
POL(p(x₁)) = [2] + x₁   
POL(s(x₁)) = x₁   

(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(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))))

K tuples:

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

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

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

We considered the (Usable) Rules:none
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)) = x₁   
POL(a__f(x₁)) = 0   
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)) = [1] + x₁   
POL(f(x₁)) = x₁   
POL(mark(x₁)) = 0   
POL(p(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(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(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))
A__F(s(0)) → c1(A__F(a__p(s(0))))

K tuples:

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

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(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))

We considered the (Usable) Rules:none
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) = [1]   
POL(A__F(x₁)) = 0   
POL(MARK(x₁)) = x₁   
POL(a__f(x₁)) = [1] + 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)) = x₁   
POL(f(x₁)) = [1] + x₁   
POL(mark(x₁)) = x₁   
POL(p(x₁)) = [1] + x₁   
POL(s(x₁)) = [1] + x₁   

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

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c5, c8, c9, c1, c6

(15) 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)))

(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))
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)))

S tuples:

A__F(s(0)) → c1(A__F(a__p(s(0))))

K tuples:

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

Defined Rule Symbols:

a__f, a__p, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c8, c9, c1, c6, c5

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

Removed 1 trailing nodes:

MARK(f(0)) → c5(A__F(0), MARK(0))

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

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

Removed 1 trailing tuple parts

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

(21) 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))))

(22) 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))))

S tuples:

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

K tuples:

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

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

Removed 2 trailing tuple parts

(24) 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(p(z0)) → c6(MARK(z0))
MARK(cons(z0, z1)) → c8(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

(25) CdtRuleRemovalProof (UPPER BOUND(ADD(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:none
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) = [1]   
POL(A__F(x₁)) = [1]   
POL(MARK(x₁)) = x₁   
POL(a__f(x₁)) = [1]   
POL(a__p(x₁)) = x₁   
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)) = [1] + x₁   
POL(f(x₁)) = [1] + x₁   
POL(mark(x₁)) = [1] + x₁   
POL(p(x₁)) = x₁   
POL(s(x₁)) = x₁   

(26) 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(p(z0)) → c6(MARK(z0))
MARK(cons(z0, z1)) → c8(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

(27) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(28) BOUNDS(1, 1)