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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs.


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


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

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

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


Rules:


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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 3
FwdInst
             ...
               →DP Problem 4
Argument Filtering and Ordering
       →DP Problem 2
FwdInst


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
INTLIST(x1) -> INTLIST(x1)
cons(x1, x2) -> cons(x1, x2)


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


Dependency Pair:


Rules:


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


Rules:


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


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


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


intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(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
Argument Filtering and 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:


intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(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 that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
INT(x1, x2) -> x2
s(x1) -> s(x1)


   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:


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


intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(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
Argument Filtering and Ordering


Dependency Pair:

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


Rules:


intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(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 that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
INT(x1, x2) -> INT(x1, x2)
s(x1) -> s(x1)


   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:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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