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.

`   R`
`     ↳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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

Dependency Pairs:

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

As we are in the innermost case, we can delete all 24 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`           →DP Problem 9`
`             ↳Size-Change Principle`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

Dependency Pairs:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. S(mark(X)) -> S(X)
2. S(ok(X)) -> S(X)
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2} , {1, 2}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
mark(x1) -> mark(x1)
ok(x1) -> ok(x1)

We obtain no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

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

As we are in the innermost case, we can delete all 24 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 10`
`             ↳Size-Change Principle`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. LENGTH1(ok(X)) -> LENGTH1(X)
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
ok(x1) -> ok(x1)

We obtain no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

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

As we are in the innermost case, we can delete all 24 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`           →DP Problem 11`
`             ↳Size-Change Principle`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

Dependency Pairs:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. FROM(ok(X)) -> FROM(X)
2. FROM(mark(X)) -> FROM(X)
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2} , {1, 2}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
mark(x1) -> mark(x1)
ok(x1) -> ok(x1)

We obtain no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

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

As we are in the innermost case, we can delete all 24 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`           →DP Problem 12`
`             ↳Size-Change Principle`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

Dependency Pairs:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
2. CONS(mark(X1), X2) -> CONS(X1, X2)
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1>1
2>2
{1, 2} , {1, 2}
1>1
2=2

which lead(s) to this/these maximal multigraph(s):
{1, 2} , {1, 2}
1>1
2=2
{1, 2} , {1, 2}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
mark(x1) -> mark(x1)
ok(x1) -> ok(x1)

We obtain no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

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

As we are in the innermost case, we can delete all 24 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`           →DP Problem 13`
`             ↳Size-Change Principle`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. LENGTH(ok(X)) -> LENGTH(X)
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
ok(x1) -> ok(x1)

We obtain no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

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

As we are in the innermost case, we can delete all 24 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`           →DP Problem 14`
`             ↳Size-Change Principle`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`

Dependency Pairs:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. ACTIVE(s(X)) -> ACTIVE(X)
2. ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
3. ACTIVE(from(X)) -> ACTIVE(X)
and get the following Size-Change Graph(s):
{1, 2, 3} , {1, 2, 3}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2, 3} , {1, 2, 3}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
from(x1) -> from(x1)
cons(x1, x2) -> cons(x1, x2)
s(x1) -> s(x1)

We obtain no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 8`
`         ↳UsableRules`

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

As we are in the innermost case, we can delete all 24 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`           →DP Problem 15`
`             ↳Size-Change Principle`
`       →DP Problem 8`
`         ↳UsableRules`

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)

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. PROPER(length1(X)) -> PROPER(X)
2. PROPER(length(X)) -> PROPER(X)
3. PROPER(s(X)) -> PROPER(X)
4. PROPER(cons(X1, X2)) -> PROPER(X2)
5. PROPER(cons(X1, X2)) -> PROPER(X1)
6. PROPER(from(X)) -> PROPER(X)
and get the following Size-Change Graph(s):
{1, 2, 3, 4, 5, 6} , {1, 2, 3, 4, 5, 6}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2, 3, 4, 5, 6} , {1, 2, 3, 4, 5, 6}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
from(x1) -> from(x1)
cons(x1, x2) -> cons(x1, x2)
s(x1) -> s(x1)
length(x1) -> length(x1)
length1(x1) -> length1(x1)

We obtain no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳Usable Rules (Innermost)`

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

As we are in the innermost case, we can delete all 2 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`
`           →DP Problem 16`
`             ↳Narrowing Transformation`

Dependency Pairs:

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

Rules:

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`
`           →DP Problem 16`
`             ↳Nar`
`             ...`
`               →DP Problem 17`
`                 ↳Narrowing Transformation`

Dependency Pairs:

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

Rules:

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`
`           →DP Problem 16`
`             ↳Nar`
`             ...`
`               →DP Problem 18`
`                 ↳Negative Polynomial Order`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x1) ) = x1

POL( ok(x1) ) = x1

POL( from(x1) ) = 1

POL( mark(x1) ) = x1

POL( cons(x1, x2) ) = 0

POL( length(x1) ) = 0

POL( s(x1) ) = 0

POL( length1(x1) ) = 0

POL( proper(x1) ) = 1

POL( nil ) = 0

POL( 0 ) = 0

POL( active(x1) ) = 1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`
`           →DP Problem 16`
`             ↳Nar`
`             ...`
`               →DP Problem 19`
`                 ↳Negative Polynomial Order`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

TOP(ok(length(cons(X'', Y')))) -> TOP(mark(s(length1(Y'))))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x1) ) = x1

POL( ok(x1) ) = x1

POL( length(x1) ) = 1

POL( mark(x1) ) = x1

POL( s(x1) ) = 0

POL( cons(x1, x2) ) = 0

POL( from(x1) ) = 0

POL( length1(x1) ) = 1

POL( proper(x1) ) = 1

POL( nil ) = 0

POL( 0 ) = 0

POL( active(x1) ) = 1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`
`           →DP Problem 16`
`             ↳Nar`
`             ...`
`               →DP Problem 20`
`                 ↳Negative Polynomial Order`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

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

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x1) ) = x1

POL( mark(x1) ) = 1

POL( length(x1) ) = 0

POL( s(x1) ) = 1

POL( cons(x1, x2) ) = 1

POL( ok(x1) ) = x1

POL( from(x1) ) = 1

POL( length1(x1) ) = 1

POL( proper(x1) ) = 1

POL( nil ) = 0

POL( 0 ) = 0

POL( active(x1) ) = 1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`
`           →DP Problem 16`
`             ↳Nar`
`             ...`
`               →DP Problem 21`
`                 ↳Dependency Graph`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`
`           →DP Problem 16`
`             ↳Nar`
`             ...`
`               →DP Problem 22`
`                 ↳Negative Polynomial Order`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

TOP(mark(length1(X''))) -> TOP(length1(proper(X'')))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x1) ) = x1

POL( mark(x1) ) = 1

POL( length1(x1) ) = 0

POL( s(x1) ) = 1

POL( cons(x1, x2) ) = 1

POL( ok(x1) ) = x1

POL( from(x1) ) = 1

POL( proper(x1) ) = 1

POL( nil ) = 0

POL( length(x1) ) = 0

POL( 0 ) = 0

POL( active(x1) ) = 1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`       →DP Problem 5`
`         ↳UsableRules`
`       →DP Problem 6`
`         ↳UsableRules`
`       →DP Problem 7`
`         ↳UsableRules`
`       →DP Problem 8`
`         ↳UsableRules`
`           →DP Problem 16`
`             ↳Nar`
`             ...`
`               →DP Problem 23`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The Proof could not be continued due to a Timeout.
Innermost Termination of R could not be shown.
Duration:
1:00 minutes