Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2]
active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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(length(cons(X, Y))) -> S(length1(Y))
ACTIVE(length(cons(X, Y))) -> LENGTH1(Y)
ACTIVE(length1(X)) -> LENGTH(X)
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)
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)
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(length(X)) -> LENGTH(proper(X))
PROPER(length(X)) -> PROPER(X)
PROPER(length1(X)) -> LENGTH1(proper(X))
PROPER(length1(X)) -> PROPER(X)
LENGTH(ok(X)) -> LENGTH(X)
LENGTH1(ok(X)) -> LENGTH1(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 eight 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

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pairs:

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

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

FROM(x₁) -> FROM(x₁)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 9

             ↳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

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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

         ↳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

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pairs:

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

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

S(x₁) -> S(x₁)
ok(x₁) -> ok(x₁)
mark(x₁) -> mark(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 10

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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

         ↳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

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pair:

LENGTH1(ok(X)) -> LENGTH1(X)

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LENGTH1(ok(X)) -> LENGTH1(X)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

LENGTH1(x₁) -> LENGTH1(x₁)
ok(x₁) -> ok(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 11

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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

         ↳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

         ↳AFS

       →DP Problem 8

         ↳Nar

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(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

CONS(x₁, x₂) -> CONS(x₁, x₂)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 12

             ↳Dependency Graph

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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

         ↳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

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pair:

LENGTH(ok(X)) -> LENGTH(X)

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LENGTH(ok(X)) -> LENGTH(X)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

LENGTH(x₁) -> LENGTH(x₁)
ok(x₁) -> ok(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

           →DP Problem 13

             ↳Dependency Graph

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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

         ↳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

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pairs:

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(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

ACTIVE(x₁) -> ACTIVE(x₁)
s(x₁) -> s(x₁)
from(x₁) -> from(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳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 14

             ↳Dependency Graph

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Nar

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Argument Filtering and Ordering

       →DP Problem 8

         ↳Nar

Dependency Pairs:

PROPER(length1(X)) -> PROPER(X)
PROPER(length(X)) -> PROPER(X)
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(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

PROPER(length1(X)) -> PROPER(X)
PROPER(length(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(from(X)) -> PROPER(X)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

PROPER(x₁) -> PROPER(x₁)
length(x₁) -> length(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
length1(x₁) -> length1(x₁)
from(x₁) -> from(x₁)
s(x₁) -> s(x₁)

     ↳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

         ↳AFS

           →DP Problem 15

             ↳Dependency Graph

       →DP Problem 8

         ↳Nar

Dependency Pair:

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Narrowing Transformation

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(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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

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

seven new Dependency Pairs are created:

TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(length(X''))) -> TOP(length(proper(X'')))
TOP(mark(nil)) -> TOP(ok(nil))
TOP(mark(0)) -> TOP(ok(0))
TOP(mark(length1(X''))) -> TOP(length1(proper(X'')))

The transformation is resulting in one new DP problem:

     ↳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

         ↳AFS

       →DP Problem 8

         ↳Nar

           →DP Problem 16

             ↳Narrowing Transformation

Dependency Pairs:

TOP(mark(length1(X''))) -> TOP(length1(proper(X'')))
TOP(mark(0)) -> TOP(ok(0))
TOP(mark(nil)) -> TOP(ok(nil))
TOP(mark(length(X''))) -> TOP(length(proper(X'')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(ok(X)) -> TOP(active(X))

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(ok(length(nil))) -> TOP(mark(0))
TOP(ok(length(cons(X'', Y')))) -> TOP(mark(s(length1(Y'))))
TOP(ok(length1(X''))) -> TOP(mark(length(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(s(X''))) -> TOP(s(active(X'')))

The transformation is resulting in one new DP problem:

     ↳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

         ↳AFS

       →DP Problem 8

         ↳Nar

           →DP Problem 16

             ↳Nar

...

               →DP Problem 17

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(length1(X''))) -> TOP(mark(length(X'')))
TOP(ok(length(cons(X'', Y')))) -> TOP(mark(s(length1(Y'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(length(X''))) -> TOP(length(proper(X'')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(length1(X''))) -> TOP(length1(proper(X'')))

Rules:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:17 minutes