Term Rewriting System R:
[y, x]
int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

INT(0, s(y)) -> INT(s(0), s(y))
INT(s(x), s(y)) -> INTLIST(int(x, y))
INT(s(x), s(y)) -> INT(x, y)
INTLIST(.(x, y)) -> INTLIST(y)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

INTLIST(.(x, y)) -> INTLIST(y)


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




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

INTLIST(.(x, y)) -> INTLIST(y)
one new Dependency Pair is created:

INTLIST(.(x, .(x'', y''))) -> INTLIST(.(x'', y''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

INTLIST(.(x, .(x'', y''))) -> INTLIST(.(x'', y''))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




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

INTLIST(.(x, .(x'', y''))) -> INTLIST(.(x'', y''))
one new Dependency Pair is created:

INTLIST(.(x, .(x'''', .(x''''', y'''')))) -> INTLIST(.(x'''', .(x''''', y'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

INTLIST(.(x, .(x'''', .(x''''', y'''')))) -> INTLIST(.(x'''', .(x''''', y'''')))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

INTLIST(.(x, .(x'''', .(x''''', y'''')))) -> INTLIST(.(x'''', .(x''''', y'''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(INT_LIST(x1))=  1 + x1  
  POL(.(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst


Dependency Pair:


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Forward Instantiation Transformation


Dependency Pairs:

INT(s(x), s(y)) -> INT(x, y)
INT(0, s(y)) -> INT(s(0), s(y))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




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

INT(s(x), s(y)) -> INT(x, y)
two new Dependency Pairs are created:

INT(s(s(x'')), s(s(y''))) -> INT(s(x''), s(y''))
INT(s(0), s(s(y''))) -> INT(0, s(y''))

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳Forward Instantiation Transformation
           →DP Problem 7
             ↳FwdInst


Dependency Pairs:

INT(s(0), s(s(y''))) -> INT(0, s(y''))
INT(0, s(y)) -> INT(s(0), s(y))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




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

INT(0, s(y)) -> INT(s(0), s(y))
one new Dependency Pair is created:

INT(0, s(s(y''''))) -> INT(s(0), s(s(y'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 8
                 ↳Forward Instantiation Transformation
           →DP Problem 7
             ↳FwdInst


Dependency Pairs:

INT(0, s(s(y''''))) -> INT(s(0), s(s(y'''')))
INT(s(0), s(s(y''))) -> INT(0, s(y''))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




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

INT(s(0), s(s(y''))) -> INT(0, s(y''))
one new Dependency Pair is created:

INT(s(0), s(s(s(y'''''')))) -> INT(0, s(s(y'''''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 10
                 ↳Forward Instantiation Transformation
           →DP Problem 7
             ↳FwdInst


Dependency Pairs:

INT(s(0), s(s(s(y'''''')))) -> INT(0, s(s(y'''''')))
INT(0, s(s(y''''))) -> INT(s(0), s(s(y'''')))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




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

INT(0, s(s(y''''))) -> INT(s(0), s(s(y'''')))
one new Dependency Pair is created:

INT(0, s(s(s(y'''''''')))) -> INT(s(0), s(s(s(y''''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 11
                 ↳Polynomial Ordering
           →DP Problem 7
             ↳FwdInst


Dependency Pairs:

INT(0, s(s(s(y'''''''')))) -> INT(s(0), s(s(s(y''''''''))))
INT(s(0), s(s(s(y'''''')))) -> INT(0, s(s(y'''''')))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

INT(s(0), s(s(s(y'''''')))) -> INT(0, s(s(y'''''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(INT(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 13
                 ↳Dependency Graph
           →DP Problem 7
             ↳FwdInst


Dependency Pair:

INT(0, s(s(s(y'''''''')))) -> INT(s(0), s(s(s(y''''''''))))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
           →DP Problem 7
             ↳Forward Instantiation Transformation


Dependency Pair:

INT(s(s(x'')), s(s(y''))) -> INT(s(x''), s(y''))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




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

INT(s(s(x'')), s(s(y''))) -> INT(s(x''), s(y''))
one new Dependency Pair is created:

INT(s(s(s(x''''))), s(s(s(y'''')))) -> INT(s(s(x'''')), s(s(y'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
           →DP Problem 7
             ↳FwdInst
             ...
               →DP Problem 9
                 ↳Polynomial Ordering


Dependency Pair:

INT(s(s(s(x''''))), s(s(s(y'''')))) -> INT(s(s(x'''')), s(s(y'''')))


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

INT(s(s(s(x''''))), s(s(s(y'''')))) -> INT(s(s(x'''')), s(s(y'''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(INT(x1, x2))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
           →DP Problem 7
             ↳FwdInst
             ...
               →DP Problem 12
                 ↳Dependency Graph


Dependency Pair:


Rules:


int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes