Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z, X1, X2]
active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
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(from(X)) -> CONS(X, from(s(X)))
ACTIVE(from(X)) -> FROM(s(X))
ACTIVE(from(X)) -> S(X)
ACTIVE(sel(s(X), cons(Y, Z))) -> SEL(X, Z)
ACTIVE(from(X)) -> FROM(active(X))
ACTIVE(from(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(sel(X1, X2)) -> SEL(active(X1), X2)
ACTIVE(sel(X1, X2)) -> ACTIVE(X1)
ACTIVE(sel(X1, X2)) -> SEL(X1, active(X2))
ACTIVE(sel(X1, X2)) -> ACTIVE(X2)
FROM(mark(X)) -> FROM(X)
FROM(ok(X)) -> FROM(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)
SEL(mark(X1), X2) -> SEL(X1, X2)
SEL(X1, mark(X2)) -> SEL(X1, X2)
SEL(ok(X1), ok(X2)) -> SEL(X1, X2)
PROPER(from(X)) -> FROM(proper(X))
PROPER(from(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(sel(X1, X2)) -> SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) -> PROPER(X1)
PROPER(sel(X1, X2)) -> PROPER(X2)
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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

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

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

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

The following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

proper > {active, sel, s} > cons > {mark, ok, from}

resulting in one new DP problem.
Used Argument Filtering System:

CONS(x₁, x₂) -> CONS(x₁, x₂)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)
active(x₁) -> active(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
from(x₁) -> from(x₁)
s(x₁) -> s(x₁)
sel(x₁, x₂) -> sel(x₁, x₂)
proper(x₁) -> proper(x₁)
top(x₁) -> top

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
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

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

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

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

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

The following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{active, cons, proper} > sel > {mark, ok, from, s}

resulting in one new DP problem.
Used Argument Filtering System:

FROM(x₁) -> FROM(x₁)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)
active(x₁) -> active(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
from(x₁) -> from(x₁)
s(x₁) -> s(x₁)
sel(x₁, x₂) -> sel(x₁, x₂)
proper(x₁) -> proper(x₁)
top(x₁) -> top

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

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

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

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

The following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

active > {proper, sel} > cons > {ok, from} > mark
active > {proper, sel} > s > {ok, from} > mark

resulting in one new DP problem.
Used Argument Filtering System:

S(x₁) -> S(x₁)
ok(x₁) -> ok(x₁)
mark(x₁) -> mark(x₁)
active(x₁) -> active(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
from(x₁) -> from(x₁)
s(x₁) -> s(x₁)
sel(x₁, x₂) -> sel(x₁, x₂)
proper(x₁) -> proper(x₁)
top(x₁) -> top

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 10

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and Ordering

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

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

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

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

The following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

proper > {active, sel, s} > cons > {ok, from} > mark

resulting in one new DP problem.
Used Argument Filtering System:

SEL(x₁, x₂) -> SEL(x₁, x₂)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)
active(x₁) -> active(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
from(x₁) -> from(x₁)
s(x₁) -> s(x₁)
sel(x₁, x₂) -> sel(x₁, x₂)
proper(x₁) -> proper(x₁)
top(x₁) -> top

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 11

             ↳Dependency Graph

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Argument Filtering and Ordering

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

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

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

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

The following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

active > {proper, sel} > cons > {mark, ok, s}
active > {proper, sel} > from > {mark, ok, s}

resulting in one new DP problem.
Used Argument Filtering System:

ACTIVE(x₁) -> ACTIVE(x₁)
s(x₁) -> s(x₁)
from(x₁) -> from(x₁)
sel(x₁, x₂) -> sel(x₁, x₂)
cons(x₁, x₂) -> cons(x₁, x₂)
active(x₁) -> active(x₁)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)
proper(x₁) -> proper(x₁)
top(x₁) -> top

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

           →DP Problem 12

             ↳Dependency Graph

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Argument Filtering and Ordering

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

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

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

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

The following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

active > {proper, from} > cons > mark
active > {proper, from} > cons > ok
active > {proper, from} > sel > mark
active > {proper, from} > sel > ok
active > {proper, from} > s > mark
active > {proper, from} > s > ok

resulting in one new DP problem.
Used Argument Filtering System:

PROPER(x₁) -> PROPER(x₁)
sel(x₁, x₂) -> sel(x₁, x₂)
cons(x₁, x₂) -> cons(x₁, x₂)
from(x₁) -> from(x₁)
s(x₁) -> s(x₁)
active(x₁) -> active(x₁)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)
proper(x₁) -> proper(x₁)
top(x₁) -> top

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

           →DP Problem 13

             ↳Dependency Graph

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
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))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
proper(0) -> ok(0)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Termination of R could not be shown.
Duration:
0:39 minutes