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

Innermost Termination of R to be shown. 



   R
     ↳Removing Redundant Rules



Removing the following rules from R which fullfill a polynomial ordering:

int(0, 0) -> cons(0, nil)
int(s(x), 0) -> nil

where the Polynomial interpretation:
  POL(intlist(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(int(x1, x2))=  1 + x1 + x2  
  POL(nil)=  0  
  POL(s(x1))=  x1  
was used.

Not all Rules of R can be deleted, so we still have to regard a part of R. 


   R
     ↳RRRPolo
       →TRS2
         ↳Removing Redundant Rules



Removing the following rules from R which fullfill a polynomial ordering:

intlist(nil) -> nil

where the Polynomial interpretation:
  POL(intlist(x1))=  2·x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(int(x1, x2))=  x1 + x2  
  POL(nil)=  1  
  POL(s(x1))=  2·x1  
was used.

Not all Rules of R can be deleted, so we still have to regard a part of R. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 1
                 ↳Usable Rules (Innermost)


Dependency Pair:

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


Rules:


int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))
intlist(cons(x, y)) -> cons(s(x), intlist(y))


Strategy:

innermost




As we are in the innermost case, we can delete all 3 non-usable-rules.


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 3
                 ↳Size-Change Principle


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. INTLIST(cons(x, y)) -> INTLIST(y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1

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

We obtain no new DP problems. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 2
                 ↳Usable Rules (Innermost)


Dependency Pairs:

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


Rules:


int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))
intlist(cons(x, y)) -> cons(s(x), intlist(y))


Strategy:

innermost




As we are in the innermost case, we can delete all 3 non-usable-rules.


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 4
                 ↳Size-Change Principle


Dependency Pairs:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. INT(0, s(y)) -> INT(s(0), s(y))
  2. INT(s(x), s(y)) -> INT(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
2=2
{2} , {2}
1>1
2>2

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

DP: empty set
Oriented Rules: none

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

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.

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