Term Rewriting System R:
[x]
top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

TOP(free(x)) -> TOP(check(new(x)))
TOP(free(x)) -> CHECK(new(x))
TOP(free(x)) -> NEW(x)
NEW(free(x)) -> NEW(x)
OLD(free(x)) -> OLD(x)
CHECK(free(x)) -> CHECK(x)
CHECK(new(x)) -> NEW(check(x))
CHECK(new(x)) -> CHECK(x)
CHECK(old(x)) -> OLD(check(x))
CHECK(old(x)) -> CHECK(x)

Furthermore, R contains four SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining


Dependency Pair:

NEW(free(x)) -> NEW(x)


Rules:


top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)





The following dependency pair can be strictly oriented:

NEW(free(x)) -> NEW(x)


Additionally, the following rules can be oriented:

top(free(x)) -> top(check(new(x)))
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(serve)=  0  
  POL(old(x1))=  1 + x1  
  POL(free(x1))=  1 + x1  
  POL(check(x1))=  x1  
  POL(NEW(x1))=  x1  
  POL(new(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining


Dependency Pair:

OLD(free(x)) -> OLD(x)


Rules:


top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)





The following dependency pair can be strictly oriented:

OLD(free(x)) -> OLD(x)


Additionally, the following rules can be oriented:

top(free(x)) -> top(check(new(x)))
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(serve)=  0  
  POL(old(x1))=  1 + x1  
  POL(free(x1))=  1 + x1  
  POL(check(x1))=  x1  
  POL(OLD(x1))=  x1  
  POL(new(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 6
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polynomial Ordering
       →DP Problem 4
         ↳Remaining


Dependency Pairs:

CHECK(old(x)) -> CHECK(x)
CHECK(new(x)) -> CHECK(x)
CHECK(free(x)) -> CHECK(x)


Rules:


top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)





The following dependency pair can be strictly oriented:

CHECK(old(x)) -> CHECK(x)


Additionally, the following rules can be oriented:

top(free(x)) -> top(check(new(x)))
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(serve)=  0  
  POL(old(x1))=  1 + x1  
  POL(free(x1))=  x1  
  POL(check(x1))=  x1  
  POL(CHECK(x1))=  x1  
  POL(new(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 7
             ↳Polynomial Ordering
       →DP Problem 4
         ↳Remaining


Dependency Pairs:

CHECK(new(x)) -> CHECK(x)
CHECK(free(x)) -> CHECK(x)


Rules:


top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)





The following dependency pair can be strictly oriented:

CHECK(new(x)) -> CHECK(x)


Additionally, the following rules can be oriented:

top(free(x)) -> top(check(new(x)))
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(serve)=  0  
  POL(old(x1))=  0  
  POL(free(x1))=  x1  
  POL(check(x1))=  x1  
  POL(CHECK(x1))=  x1  
  POL(new(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 7
             ↳Polo
             ...
               →DP Problem 8
                 ↳Polynomial Ordering
       →DP Problem 4
         ↳Remaining


Dependency Pair:

CHECK(free(x)) -> CHECK(x)


Rules:


top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)





The following dependency pair can be strictly oriented:

CHECK(free(x)) -> CHECK(x)


Additionally, the following rules can be oriented:

top(free(x)) -> top(check(new(x)))
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(serve)=  0  
  POL(old(x1))=  1 + x1  
  POL(free(x1))=  1 + x1  
  POL(check(x1))=  x1  
  POL(CHECK(x1))=  x1  
  POL(new(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 7
             ↳Polo
             ...
               →DP Problem 9
                 ↳Dependency Graph
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

TOP(free(x)) -> TOP(check(new(x)))


Rules:


top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)




Termination of R could not be shown.
Duration:
0:00 minutes