Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z, X1, X2]
active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(from(X)) -> CONS(X, from(s(X)))
ACTIVE(from(X)) -> FROM(s(X))
ACTIVE(from(X)) -> S(X)
ACTIVE(2nd(X)) -> 2ND(active(X))
ACTIVE(2nd(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(from(X)) -> FROM(active(X))
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
2ND(mark(X)) -> 2ND(X)
2ND(ok(X)) -> 2ND(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
FROM(mark(X)) -> FROM(X)
FROM(ok(X)) -> FROM(X)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
PROPER(2nd(X)) -> 2ND(proper(X))
PROPER(2nd(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(from(X)) -> FROM(proper(X))
PROPER(from(X)) -> PROPER(X)
PROPER(s(X)) -> S(proper(X))
PROPER(s(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

         ↳Polo

Dependency Pairs:

FROM(ok(X)) -> FROM(X)
FROM(mark(X)) -> FROM(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

FROM(ok(X)) -> FROM(X)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(FROM(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(ok(x₁)) = 1 + 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

         ↳Polo

Dependency Pair:

FROM(mark(X)) -> FROM(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

FROM(mark(X)) -> FROM(X)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(FROM(x₁)) = x₁

POL(mark(x₁)) = 1 + 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

         ↳Polo

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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

         ↳Polo

Dependency Pairs:

2ND(ok(X)) -> 2ND(X)
2ND(mark(X)) -> 2ND(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

2ND(ok(X)) -> 2ND(X)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(2ND(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

         ↳Polo

Dependency Pair:

2ND(mark(X)) -> 2ND(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

2ND(mark(X)) -> 2ND(X)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(2ND(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

         ↳Polo

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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

         ↳Polo

Dependency Pairs:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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

         ↳Polo

Dependency Pair:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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

         ↳Polo

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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

         ↳Polo

Dependency Pairs:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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

         ↳Polo

Dependency Pair:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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

         ↳Polo

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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

         ↳Polo

Dependency Pairs:

ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(2nd(X)) -> ACTIVE(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVE(2nd(X)) -> ACTIVE(X)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

POL(ACTIVE(x₁)) = x₁

POL(2nd(x₁)) = 1 + x₁

POL(cons(x₁, x₂)) = x₁

POL(s(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

         ↳Polo

Dependency Pairs:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

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 17

                 ↳Polynomial Ordering

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pairs:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVE(from(X)) -> ACTIVE(X)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(ACTIVE(x₁)) = 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 4

         ↳Polo

       →DP Problem 5

         ↳Polo

           →DP Problem 16

             ↳Polo

...

               →DP Problem 18

                 ↳Polynomial Ordering

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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

         ↳Polo

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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

         ↳Polo

Dependency Pairs:

PROPER(s(X)) -> PROPER(X)
PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(2nd(X)) -> PROPER(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

PROPER(2nd(X)) -> PROPER(X)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

POL(2nd(x₁)) = 1 + x₁

POL(PROPER(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁ + x₂

POL(s(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

         ↳Polo

Dependency Pairs:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

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 21

                 ↳Polynomial Ordering

       →DP Problem 7

         ↳Polo

Dependency Pairs:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

PROPER(from(X)) -> PROPER(X)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(PROPER(x₁)) = x₁

POL(cons(x₁, x₂)) = 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 22

                 ↳Polynomial Ordering

       →DP Problem 7

         ↳Polo

Dependency Pairs:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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 for innermost 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

         ↳Polo

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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

         ↳Polynomial Ordering

Dependency Pairs:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(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))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

POL(active(x₁)) = 0

POL(proper(x₁)) = 0

POL(2nd(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁

POL(s(x₁)) = x₁

POL(mark(x₁)) = 0

POL(ok(x₁)) = 1

POL(TOP(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 7

         ↳Polo

           →DP Problem 24

             ↳Polynomial Ordering

Dependency Pair:

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

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

POL(proper(x₁)) = 0

POL(2nd(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁

POL(s(x₁)) = x₁

POL(mark(x₁)) = 1

POL(ok(x₁)) = 0

POL(TOP(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 7

         ↳Polo

           →DP Problem 24

             ↳Polo

...

               →DP Problem 25

                 ↳Dependency Graph

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes

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

POL(2ND(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(from(x₁))	= x₁
POL(ACTIVE(x₁))	= x₁
POL(2nd(x₁))	= 1 + x₁
POL(cons(x₁, x₂))	= x₁
POL(s(x₁))	= x₁

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