Termination Proof using AProVE

Term Rewriting System R:
[X, X1, X2]
active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(f(0)) -> CONS(0, f(s(0)))
ACTIVE(f(0)) -> F(s(0))
ACTIVE(f(0)) -> S(0)
ACTIVE(f(s(0))) -> F(p(s(0)))
ACTIVE(f(s(0))) -> P(s(0))
ACTIVE(f(X)) -> F(active(X))
ACTIVE(f(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(p(X)) -> P(active(X))
ACTIVE(p(X)) -> ACTIVE(X)
F(mark(X)) -> F(X)
F(ok(X)) -> F(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
P(mark(X)) -> P(X)
P(ok(X)) -> P(X)
PROPER(f(X)) -> F(proper(X))
PROPER(f(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(p(X)) -> P(proper(X))
PROPER(p(X)) -> PROPER(X)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains seven SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pairs:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

F(ok(X)) -> F(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(mark(x₁)) = x₁

POL(ok(x₁)) = 1 + x₁

POL(F(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 8

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

F(mark(X)) -> F(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

F(mark(X)) -> F(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(mark(x₁)) = 1 + x₁

POL(F(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 8

             ↳Polo

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pairs:

P(ok(X)) -> P(X)
P(mark(X)) -> P(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

P(ok(X)) -> P(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(P(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(ok(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 10

             ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

P(mark(X)) -> P(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

P(mark(X)) -> P(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(P(x₁)) = x₁

POL(mark(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 10

             ↳Polo

...

               →DP Problem 11

                 ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pairs:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(mark(x₁)) = 0

POL(ok(x₁)) = 1 + x₁

POL(CONS(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 12

             ↳Polynomial Ordering

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

CONS(mark(X1), X2) -> CONS(X1, X2)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

CONS(mark(X1), X2) -> CONS(X1, X2)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(mark(x₁)) = 1 + x₁

POL(CONS(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 12

             ↳Polo

...

               →DP Problem 13

                 ↳Dependency Graph

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polynomial Ordering

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pairs:

S(ok(X)) -> S(X)
S(mark(X)) -> S(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

S(ok(X)) -> S(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(S(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(ok(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

           →DP Problem 14

             ↳Polynomial Ordering

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

S(mark(X)) -> S(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

S(mark(X)) -> S(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(S(x₁)) = x₁

POL(mark(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

           →DP Problem 14

             ↳Polo

...

               →DP Problem 15

                 ↳Dependency Graph

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polynomial Ordering

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pairs:

ACTIVE(p(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(f(X)) -> ACTIVE(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(ACTIVE(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁

POL(s(x₁)) = x₁

POL(f(x₁)) = 1 + x₁

POL(p(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

           →DP Problem 16

             ↳Polynomial Ordering

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pairs:

ACTIVE(p(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(p(X)) -> ACTIVE(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(ACTIVE(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁

POL(s(x₁)) = x₁

POL(p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

           →DP Problem 16

             ↳Polo

...

               →DP Problem 17

                 ↳Polynomial Ordering

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pairs:

ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(s(X)) -> ACTIVE(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(ACTIVE(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

           →DP Problem 16

             ↳Polo

...

               →DP Problem 18

                 ↳Polynomial Ordering

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(ACTIVE(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

           →DP Problem 16

             ↳Polo

...

               →DP Problem 19

                 ↳Dependency Graph

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polynomial Ordering

       →DP Problem 7

         ↳Nar

Dependency Pairs:

PROPER(p(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

PROPER(f(X)) -> PROPER(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(PROPER(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = x₁

POL(f(x₁)) = 1 + x₁

POL(p(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

           →DP Problem 20

             ↳Polynomial Ordering

       →DP Problem 7

         ↳Nar

Dependency Pairs:

PROPER(p(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

PROPER(p(X)) -> PROPER(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(PROPER(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = x₁

POL(p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

           →DP Problem 20

             ↳Polo

...

               →DP Problem 21

                 ↳Polynomial Ordering

       →DP Problem 7

         ↳Nar

Dependency Pairs:

PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

PROPER(s(X)) -> PROPER(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(PROPER(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

           →DP Problem 20

             ↳Polo

...

               →DP Problem 22

                 ↳Polynomial Ordering

       →DP Problem 7

         ↳Nar

Dependency Pairs:

PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(PROPER(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

           →DP Problem 20

             ↳Polo

...

               →DP Problem 23

                 ↳Dependency Graph

       →DP Problem 7

         ↳Nar

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Narrowing Transformation

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(X)) -> TOP(proper(X))

five new Dependency Pairs are created:

TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(mark(0)) -> TOP(ok(0))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(p(X''))) -> TOP(p(proper(X'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

           →DP Problem 24

             ↳Narrowing Transformation

Dependency Pairs:

TOP(mark(p(X''))) -> TOP(p(proper(X'')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(0)) -> TOP(ok(0))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(ok(X)) -> TOP(active(X))

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(X)) -> TOP(active(X))

seven new Dependency Pairs are created:

TOP(ok(f(0))) -> TOP(mark(cons(0, f(s(0)))))
TOP(ok(f(s(0)))) -> TOP(mark(f(p(s(0)))))
TOP(ok(p(s(0)))) -> TOP(mark(0))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(p(X''))) -> TOP(p(active(X'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

           →DP Problem 24

             ↳Nar

...

               →DP Problem 25

                 ↳Polynomial Ordering

Dependency Pairs:

TOP(ok(p(X''))) -> TOP(p(active(X'')))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(f(s(0)))) -> TOP(mark(f(p(s(0)))))
TOP(ok(f(0))) -> TOP(mark(cons(0, f(s(0)))))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(mark(p(X''))) -> TOP(p(proper(X'')))

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

TOP(ok(f(0))) -> TOP(mark(cons(0, f(s(0)))))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(active(x₁)) = 0

POL(proper(x₁)) = 0

POL(0) = 0

POL(cons(x₁, x₂)) = 0

POL(s(x₁)) = 0

POL(mark(x₁)) = x₁

POL(ok(x₁)) = x₁

POL(TOP(x₁)) = x₁

POL(f(x₁)) = 1

POL(p(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

           →DP Problem 24

             ↳Nar

...

               →DP Problem 26

                 ↳Polynomial Ordering

Dependency Pairs:

TOP(ok(p(X''))) -> TOP(p(active(X'')))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(f(s(0)))) -> TOP(mark(f(p(s(0)))))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(mark(p(X''))) -> TOP(p(proper(X'')))

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

TOP(ok(f(s(0)))) -> TOP(mark(f(p(s(0)))))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(active(x₁)) = x₁

POL(proper(x₁)) = x₁

POL(0) = 0

POL(cons(x₁, x₂)) = 0

POL(s(x₁)) = 1

POL(mark(x₁)) = x₁

POL(ok(x₁)) = x₁

POL(TOP(x₁)) = 1 + x₁

POL(f(x₁)) = x₁

POL(p(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Nar

           →DP Problem 24

             ↳Nar

...

               →DP Problem 27

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

TOP(ok(p(X''))) -> TOP(p(active(X'')))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(mark(p(X''))) -> TOP(p(proper(X'')))

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Termination of R could not be shown.
Duration:
0:20 minutes

POL(mark(x₁))	= x₁
POL(ok(x₁))	= 1 + x₁
POL(F(x₁))	= x₁

POL(P(x₁))	= x₁
POL(mark(x₁))	= x₁
POL(ok(x₁))	= 1 + x₁

POL(mark(x₁))	= 1 + x₁
POL(CONS(x₁, x₂))	= x₁

POL(S(x₁))	= x₁
POL(mark(x₁))	= x₁
POL(ok(x₁))	= 1 + x₁

POL(ACTIVE(x₁))	= x₁
POL(cons(x₁, x₂))	= x₁
POL(s(x₁))	= x₁
POL(f(x₁))	= 1 + x₁
POL(p(x₁))	= x₁

POL(PROPER(x₁))	= x₁
POL(cons(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= x₁
POL(f(x₁))	= 1 + x₁
POL(p(x₁))	= x₁

POL(active(x₁))	= 0
POL(proper(x₁))	= 0
POL(0)	= 0
POL(cons(x₁, x₂))	= 0
POL(s(x₁))	= 0
POL(mark(x₁))	= x₁
POL(ok(x₁))	= x₁
POL(TOP(x₁))	= x₁
POL(f(x₁))	= 1
POL(p(x₁))	= 0