Term Rewriting System R:
[X, X1, X2]
active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(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(f(0)) -> CONS(0, f(s(0)))
ACTIVE(f(0)) -> F(s(0))
ACTIVE(f(0)) -> S(0)
ACTIVE(f(s(0))) -> F(p(s(0)))
ACTIVE(f(s(0))) -> P(s(0))
ACTIVE(f(X)) -> F(active(X))
ACTIVE(f(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(p(X)) -> P(active(X))
ACTIVE(p(X)) -> ACTIVE(X)
F(mark(X)) -> F(X)
F(ok(X)) -> F(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)
P(mark(X)) -> P(X)
P(ok(X)) -> P(X)
PROPER(f(X)) -> F(proper(X))
PROPER(f(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(p(X)) -> P(proper(X))
PROPER(p(X)) -> PROPER(X)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains seven SCCs. 


   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


Dependency Pairs:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(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 22 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 8
             ↳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


Dependency Pairs:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)


Rule:

none


Strategy:

innermost




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


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(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 22 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 9
             ↳Size-Change Principle
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳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
         ↳Usable Rules (Innermost)
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(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 22 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
           →DP Problem 10
             ↳Size-Change Principle
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳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
         ↳Usable Rules (Innermost)
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules


Dependency Pairs:

P(ok(X)) -> P(X)
P(mark(X)) -> P(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(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 22 non-usable-rules.


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


Dependency Pairs:

P(ok(X)) -> P(X)
P(mark(X)) -> P(X)


Rule:

none


Strategy:

innermost




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


Dependency Pairs:

ACTIVE(p(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(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 22 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 12
             ↳Size-Change Principle
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules


Dependency Pairs:

ACTIVE(p(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. ACTIVE(p(X)) -> ACTIVE(X)
  2. ACTIVE(s(X)) -> ACTIVE(X)
  3. ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
  4. ACTIVE(f(X)) -> ACTIVE(X)
and get the following Size-Change Graph(s):
{1, 2, 3, 4} , {1, 2, 3, 4}
1>1

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

DP: empty set
Oriented Rules: none

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

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


Dependency Pairs:

PROPER(p(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(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 22 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 13
             ↳Size-Change Principle
       →DP Problem 7
         ↳UsableRules


Dependency Pairs:

PROPER(p(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)


Rule:

none


Strategy:

innermost




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

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

DP: empty set
Oriented Rules: none

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

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


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(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 14
             ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(f(s(0))) -> mark(f(p(s(0))))
active(p(X)) -> p(active(X))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(f(0)) -> mark(cons(0, f(s(0))))
active(s(X)) -> s(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(p(X)) -> p(proper(X))
proper(f(X)) -> f(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(f(s(0)))) -> TOP(mark(f(p(s(0)))))
TOP(ok(p(X''))) -> TOP(p(active(X'')))
TOP(ok(p(s(0)))) -> TOP(mark(0))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(f(0))) -> TOP(mark(cons(0, f(s(0)))))
TOP(ok(s(X''))) -> TOP(s(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 14
             ↳Nar
             ...
               →DP Problem 15
                 ↳Narrowing Transformation


Dependency Pairs:

TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(f(0))) -> TOP(mark(cons(0, f(s(0)))))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(p(s(0)))) -> TOP(mark(0))
TOP(ok(p(X''))) -> TOP(p(active(X'')))
TOP(ok(f(s(0)))) -> TOP(mark(f(p(s(0)))))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(mark(X)) -> TOP(proper(X))


Rules:


p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(f(s(0))) -> mark(f(p(s(0))))
active(p(X)) -> p(active(X))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(f(0)) -> mark(cons(0, f(s(0))))
active(s(X)) -> s(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(p(X)) -> p(proper(X))
proper(f(X)) -> f(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))
five new Dependency Pairs are created:

TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(p(X''))) -> TOP(p(proper(X'')))
TOP(mark(f(X''))) -> TOP(f(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 14
             ↳Nar
             ...
               →DP Problem 16
                 ↳Negative Polynomial Order


Dependency Pairs:

TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(p(X''))) -> TOP(p(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(ok(f(0))) -> TOP(mark(cons(0, f(s(0)))))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(p(X''))) -> TOP(p(active(X'')))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(ok(f(s(0)))) -> TOP(mark(f(p(s(0)))))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(s(X''))) -> TOP(s(active(X'')))


Rules:


p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(f(s(0))) -> mark(f(p(s(0))))
active(p(X)) -> p(active(X))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(f(0)) -> mark(cons(0, f(s(0))))
active(s(X)) -> s(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(p(X)) -> p(proper(X))
proper(f(X)) -> f(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(f(0))) -> TOP(mark(cons(0, f(s(0)))))


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

p(mark(X)) -> mark(p(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
f(mark(X)) -> mark(f(X))
s(ok(X)) -> ok(s(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
f(ok(X)) -> ok(f(X))
proper(f(X)) -> f(proper(X))
proper(p(X)) -> p(proper(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(f(s(0))) -> mark(f(p(s(0))))
active(p(X)) -> p(active(X))
active(p(s(0))) -> mark(0)
active(f(0)) -> mark(cons(0, f(s(0))))
active(f(X)) -> f(active(X))
active(s(X)) -> s(active(X))


Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x1) ) = x1

POL( ok(x1) ) = x1

POL( f(x1) ) = 1

POL( mark(x1) ) = x1

POL( cons(x1, x2) ) = 0

POL( s(x1) ) = 0

POL( p(x1) ) = 0

POL( proper(x1) ) = 1

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 14
             ↳Nar
             ...
               →DP Problem 17
                 ↳Negative Polynomial Order


Dependency Pairs:

TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(p(X''))) -> TOP(p(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(p(X''))) -> TOP(p(active(X'')))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(ok(f(s(0)))) -> TOP(mark(f(p(s(0)))))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(s(X''))) -> TOP(s(active(X'')))


Rules:


p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(f(s(0))) -> mark(f(p(s(0))))
active(p(X)) -> p(active(X))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(f(0)) -> mark(cons(0, f(s(0))))
active(s(X)) -> s(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(p(X)) -> p(proper(X))
proper(f(X)) -> f(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(f(s(0)))) -> TOP(mark(f(p(s(0)))))


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

p(mark(X)) -> mark(p(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
f(mark(X)) -> mark(f(X))
s(ok(X)) -> ok(s(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
f(ok(X)) -> ok(f(X))
proper(f(X)) -> f(proper(X))
proper(p(X)) -> p(proper(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(f(s(0))) -> mark(f(p(s(0))))
active(p(X)) -> p(active(X))
active(p(s(0))) -> mark(0)
active(f(0)) -> mark(cons(0, f(s(0))))
active(f(X)) -> f(active(X))
active(s(X)) -> s(active(X))


Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x1) ) = x1

POL( ok(x1) ) = x1

POL( f(x1) ) = x1

POL( s(x1) ) = 1

POL( mark(x1) ) = x1

POL( p(x1) ) = 0

POL( cons(x1, x2) ) = 0

POL( proper(x1) ) = x1

POL( active(x1) ) = x1

POL( 0 ) = 0


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 14
             ↳Nar
             ...
               →DP Problem 18
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(f(s(0))) -> mark(f(p(s(0))))
active(p(X)) -> p(active(X))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(f(0)) -> mark(cons(0, f(s(0))))
active(s(X)) -> s(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
s(ok(X)) -> ok(s(X))
s(mark(X)) -> mark(s(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(p(X)) -> p(proper(X))
proper(f(X)) -> f(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