Term Rewriting System R:
[x, y]
top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

TOP(sent(x)) -> TOP(check(rest(x)))
TOP(sent(x)) -> CHECK(rest(x))
TOP(sent(x)) -> REST(x)
CHECK(sent(x)) -> CHECK(x)
CHECK(rest(x)) -> REST(check(x))
CHECK(rest(x)) -> CHECK(x)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(cons(x, y)) -> CHECK(y)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules


Dependency Pairs:

CHECK(cons(x, y)) -> CHECK(y)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(rest(x)) -> CHECK(x)
CHECK(sent(x)) -> CHECK(x)


Rules:


top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)


Strategy:

innermost




As we are in the innermost case, we can delete all 8 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 3
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules


Dependency Pairs:

CHECK(cons(x, y)) -> CHECK(y)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(rest(x)) -> CHECK(x)
CHECK(sent(x)) -> CHECK(x)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. CHECK(cons(x, y)) -> CHECK(y)
  2. CHECK(cons(x, y)) -> CHECK(x)
  3. CHECK(rest(x)) -> CHECK(x)
  4. CHECK(sent(x)) -> CHECK(x)
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1>1
{3} , {3}
1>1
{4} , {4}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2} , {1, 2}
1>1
{4} , {4}
1>1
{3} , {3}
1>1
{4} , {3}
1>1
{4} , {1, 2}
1>1
{3} , {4}
1>1
{1, 2} , {4}
1>1
{3} , {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:
rest(x1) -> rest(x1)
cons(x1, x2) -> cons(x1, x2)
sent(x1) -> sent(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)


Dependency Pair:

TOP(sent(x)) -> TOP(check(rest(x)))


Rules:


top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)


Strategy:

innermost




As we are in the innermost case, we can delete all 1 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳Modular Removal of Rules


Dependency Pair:

TOP(sent(x)) -> TOP(check(rest(x)))


Rules:


check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
rest(cons(x, y)) -> sent(y)
rest(nil) -> sent(nil)


Strategy:

innermost




We have the following set of usable rules:

check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
rest(cons(x, y)) -> sent(y)
rest(nil) -> sent(nil)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(rest(x1))=  x1  
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(check(x1))=  x1  
  POL(nil)=  0  
  POL(TOP(x1))=  x1  
  POL(sent(x1))=  x1  

We have the following set D of usable symbols: {rest, cons, nil, check, TOP, sent}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

rest(cons(x, y)) -> sent(y)


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳MRR
             ...
               →DP Problem 5
                 ↳Narrowing Transformation


Dependency Pair:

TOP(sent(x)) -> TOP(check(rest(x)))


Rules:


check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
rest(nil) -> sent(nil)


Strategy:

innermost




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

TOP(sent(x)) -> TOP(check(rest(x)))
two new Dependency Pairs are created:

TOP(sent(x'')) -> TOP(rest(check(x'')))
TOP(sent(nil)) -> TOP(check(sent(nil)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳MRR
             ...
               →DP Problem 6
                 ↳Negative Polynomial Order


Dependency Pairs:

TOP(sent(nil)) -> TOP(check(sent(nil)))
TOP(sent(x'')) -> TOP(rest(check(x'')))


Rules:


check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
rest(nil) -> sent(nil)


Strategy:

innermost




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

TOP(sent(nil)) -> TOP(check(sent(nil)))


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

check(cons(x, y)) -> cons(x, check(y))
rest(nil) -> sent(nil)
check(cons(x, y)) -> cons(x, y)
check(rest(x)) -> rest(check(x))
check(sent(x)) -> sent(check(x))
check(cons(x, y)) -> cons(check(x), y)


Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x1) ) = x1

POL( sent(x1) ) = x1

POL( nil ) = 1

POL( check(x1) ) = 0

POL( rest(x1) ) = x1

POL( cons(x1, x2) ) = 0


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳MRR
             ...
               →DP Problem 7
                 ↳Semantic Labelling


Dependency Pair:

TOP(sent(x'')) -> TOP(rest(check(x'')))


Rules:


check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
rest(nil) -> sent(nil)


Strategy:

innermost




Using Semantic Labelling, the DP problem could be transformed. The following model was found:
check(x0)=  0
cons(x0, x1)=  0
TOP(x0)=  0
sent(x0)=  0
rest(x0)=  0
nil=  1

From the dependency graph we obtain 0 (labeled) SCCs which each result in correspondingDP problem.

Innermost Termination of R successfully shown.
Duration:
0:03 minutes