Term Rewriting System R:
[X, Y, Z, X1, X2]
active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
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(2nd(cons(X, X1))) -> 2ND(cons1(X, X1))
ACTIVE(2nd(cons(X, X1))) -> CONS1(X, X1)
ACTIVE(from(X)) -> CONS(X, from(s(X)))
ACTIVE(from(X)) -> FROM(s(X))
ACTIVE(from(X)) -> S(X)
ACTIVE(2nd(X)) -> 2ND(active(X))
ACTIVE(2nd(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(from(X)) -> FROM(active(X))
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons1(X1, X2)) -> CONS1(active(X1), X2)
ACTIVE(cons1(X1, X2)) -> ACTIVE(X1)
ACTIVE(cons1(X1, X2)) -> CONS1(X1, active(X2))
ACTIVE(cons1(X1, X2)) -> ACTIVE(X2)
2ND(mark(X)) -> 2ND(X)
2ND(ok(X)) -> 2ND(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
FROM(mark(X)) -> FROM(X)
FROM(ok(X)) -> FROM(X)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
CONS1(mark(X1), X2) -> CONS1(X1, X2)
CONS1(X1, mark(X2)) -> CONS1(X1, X2)
CONS1(ok(X1), ok(X2)) -> CONS1(X1, X2)
PROPER(2nd(X)) -> 2ND(proper(X))
PROPER(2nd(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(from(X)) -> FROM(proper(X))
PROPER(from(X)) -> PROPER(X)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(cons1(X1, X2)) -> CONS1(proper(X1), proper(X2))
PROPER(cons1(X1, X2)) -> PROPER(X1)
PROPER(cons1(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 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:

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


Rules:


active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
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 27 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:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. 2ND(ok(X)) -> 2ND(X)
  2. 2ND(mark(X)) -> 2ND(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 Pairs:

CONS1(ok(X1), ok(X2)) -> CONS1(X1, X2)
CONS1(mark(X1), X2) -> CONS1(X1, X2)
CONS1(X1, mark(X2)) -> CONS1(X1, X2)


Rules:


active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
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 27 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 Pairs:

CONS1(ok(X1), ok(X2)) -> CONS1(X1, X2)
CONS1(mark(X1), X2) -> CONS1(X1, X2)
CONS1(X1, mark(X2)) -> CONS1(X1, X2)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. CONS1(ok(X1), ok(X2)) -> CONS1(X1, X2)
  2. CONS1(mark(X1), X2) -> CONS1(X1, X2)
  3. CONS1(X1, mark(X2)) -> CONS1(X1, X2)
and get the following Size-Change Graph(s):
{1, 2, 3} , {1, 2, 3}
1>1
2>2
{1, 2, 3} , {1, 2, 3}
1>1
2=2
{1, 2, 3} , {1, 2, 3}
1=1
2>2

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

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


Rules:


active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
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 27 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:

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
         ↳Usable Rules (Innermost)
       →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(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
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 27 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:

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
         ↳UsableRules
       →DP Problem 5
         ↳Usable Rules (Innermost)
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules


Dependency Pairs:

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


Rules:


active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
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 27 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 Pairs:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. S(ok(X)) -> S(X)
  2. S(mark(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
         ↳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(cons1(X1, X2)) -> ACTIVE(X2)
ACTIVE(cons1(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(2nd(X)) -> ACTIVE(X)


Rules:


active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
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 27 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(cons1(X1, X2)) -> ACTIVE(X2)
ACTIVE(cons1(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(2nd(X)) -> ACTIVE(X)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. ACTIVE(cons1(X1, X2)) -> ACTIVE(X2)
  2. ACTIVE(cons1(X1, X2)) -> ACTIVE(X1)
  3. ACTIVE(s(X)) -> ACTIVE(X)
  4. ACTIVE(from(X)) -> ACTIVE(X)
  5. ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
  6. ACTIVE(2nd(X)) -> ACTIVE(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)
cons1(x1, x2) -> cons1(x1, x2)
2nd(x1) -> 2nd(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(cons1(X1, X2)) -> PROPER(X2)
PROPER(cons1(X1, X2)) -> PROPER(X1)
PROPER(s(X)) -> PROPER(X)
PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(2nd(X)) -> PROPER(X)


Rules:


active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
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 27 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(cons1(X1, X2)) -> PROPER(X2)
PROPER(cons1(X1, X2)) -> PROPER(X1)
PROPER(s(X)) -> PROPER(X)
PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(2nd(X)) -> PROPER(X)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. PROPER(cons1(X1, X2)) -> PROPER(X2)
  2. PROPER(cons1(X1, X2)) -> PROPER(X1)
  3. PROPER(s(X)) -> PROPER(X)
  4. PROPER(from(X)) -> PROPER(X)
  5. PROPER(cons(X1, X2)) -> PROPER(X2)
  6. PROPER(cons(X1, X2)) -> PROPER(X1)
  7. PROPER(2nd(X)) -> PROPER(X)
and get the following Size-Change Graph(s):
{1, 2, 3, 4, 5, 6, 7} , {1, 2, 3, 4, 5, 6, 7}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2, 3, 4, 5, 6, 7} , {1, 2, 3, 4, 5, 6, 7}
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)
cons1(x1, x2) -> cons1(x1, x2)
2nd(x1) -> 2nd(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
         ↳UsableRules
       →DP Problem 8
         ↳Usable Rules (Innermost)


Dependency Pairs:

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


Rules:


active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
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
             ↳Negative Polynomial Order


Dependency Pairs:

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


Rules:


active(cons(X1, X2)) -> cons(active(X1), X2)
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(cons1(X1, X2)) -> cons1(X1, active(X2))
active(2nd(X)) -> 2nd(active(X))
active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
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))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
from(ok(X)) -> ok(from(X))
from(mark(X)) -> mark(from(X))
2nd(ok(X)) -> ok(2nd(X))
2nd(mark(X)) -> mark(2nd(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
proper(2nd(X)) -> 2nd(proper(X))


Strategy:

innermost




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

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


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

active(cons(X1, X2)) -> cons(active(X1), X2)
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(cons1(X1, X2)) -> cons1(X1, active(X2))
active(2nd(X)) -> 2nd(active(X))
active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
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))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
from(ok(X)) -> ok(from(X))
from(mark(X)) -> mark(from(X))
2nd(ok(X)) -> ok(2nd(X))
2nd(mark(X)) -> mark(2nd(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
proper(2nd(X)) -> 2nd(proper(X))


Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x1) ) = x1

POL( ok(x1) ) = x1 + 1

POL( active(x1) ) = x1

POL( mark(x1) ) = 0

POL( proper(x1) ) = 0

POL( cons(x1, x2) ) = x2

POL( cons1(x1, x2) ) = x2

POL( 2nd(x1) ) = x1

POL( s(x1) ) = x1

POL( from(x1) ) = x1


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
             ↳Neg POLO
             ...
               →DP Problem 17
                 ↳Usable Rules (Innermost)


Dependency Pair:

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


Rules:


active(cons(X1, X2)) -> cons(active(X1), X2)
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(cons1(X1, X2)) -> cons1(X1, active(X2))
active(2nd(X)) -> 2nd(active(X))
active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
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))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
from(ok(X)) -> ok(from(X))
from(mark(X)) -> mark(from(X))
2nd(ok(X)) -> ok(2nd(X))
2nd(mark(X)) -> mark(2nd(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
proper(2nd(X)) -> 2nd(proper(X))


Strategy:

innermost




As we are in the innermost case, we can delete all 9 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
             ↳Neg POLO
             ...
               →DP Problem 18
                 ↳Modular Removal of Rules


Dependency Pair:

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


Rules:


cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
from(ok(X)) -> ok(from(X))
from(mark(X)) -> mark(from(X))
2nd(ok(X)) -> ok(2nd(X))
2nd(mark(X)) -> mark(2nd(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))


Strategy:

innermost




We have the following set of usable rules:

cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
from(ok(X)) -> ok(from(X))
from(mark(X)) -> mark(from(X))
2nd(ok(X)) -> ok(2nd(X))
2nd(mark(X)) -> mark(2nd(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  x1  
  POL(cons1(x1, x2))=  x1 + x2  
  POL(2nd(x1))=  x1  
  POL(cons(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(TOP(x1))=  x1  
  POL(ok(x1))=  1 + x1  

We have the following set D of usable symbols: {from, proper, cons1, 2nd, cons, s, mark, TOP, ok}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))


The result of this processor delivers 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
             ↳Neg POLO
             ...
               →DP Problem 19
                 ↳Modular Removal of Rules


Dependency Pair:

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


Rules:


cons(mark(X1), X2) -> mark(cons(X1, X2))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
from(ok(X)) -> ok(from(X))
from(mark(X)) -> mark(from(X))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
2nd(ok(X)) -> ok(2nd(X))
2nd(mark(X)) -> mark(2nd(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))


Strategy:

innermost




We have the following set of usable rules:

cons(mark(X1), X2) -> mark(cons(X1, X2))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
s(ok(X)) -> ok(s(X))
from(ok(X)) -> ok(from(X))
from(mark(X)) -> mark(from(X))
proper(from(X)) -> from(proper(X))
s(mark(X)) -> mark(s(X))
2nd(ok(X)) -> ok(2nd(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
proper(s(X)) -> s(proper(X))
2nd(mark(X)) -> mark(2nd(X))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
proper(2nd(X)) -> 2nd(proper(X))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  x1  
  POL(cons1(x1, x2))=  x1 + x2  
  POL(2nd(x1))=  x1  
  POL(cons(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  
  POL(mark(x1))=  1 + x1  
  POL(TOP(x1))=  x1  
  POL(ok(x1))=  x1  

We have the following set D of usable symbols: {from, proper, cons1, 2nd, cons, s, mark, TOP, ok}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.


Innermost Termination of R successfully shown.
Duration:
0:08 minutes