Term Rewriting System R:
[X, Z, Y, X1, X2]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains two SCCs.


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


Dependency Pairs:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




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

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
one new Dependency Pair is created:

FIRST(s(X), cons(Y, nfirst(X1'', X2''))) -> ACTIVATE(nfirst(X1'', X2''))

The transformation is resulting in one new DP problem:



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


Dependency Pairs:

FIRST(s(X), cons(Y, nfirst(X1'', X2''))) -> ACTIVATE(nfirst(X1'', X2''))
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




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

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
one new Dependency Pair is created:

ACTIVATE(nfirst(s(X''), cons(Y'', nfirst(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', nfirst(X1'''', X2'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 3
FwdInst
             ...
               →DP Problem 4
Forward Instantiation Transformation
       →DP Problem 2
Nar


Dependency Pairs:

ACTIVATE(nfirst(s(X''), cons(Y'', nfirst(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', nfirst(X1'''', X2'''')))
FIRST(s(X), cons(Y, nfirst(X1'', X2''))) -> ACTIVATE(nfirst(X1'', X2''))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




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

FIRST(s(X), cons(Y, nfirst(X1'', X2''))) -> ACTIVATE(nfirst(X1'', X2''))
one new Dependency Pair is created:

FIRST(s(X), cons(Y, nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))) -> ACTIVATE(nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 3
FwdInst
             ...
               →DP Problem 5
Forward Instantiation Transformation
       →DP Problem 2
Nar


Dependency Pairs:

FIRST(s(X), cons(Y, nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))) -> ACTIVATE(nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))
ACTIVATE(nfirst(s(X''), cons(Y'', nfirst(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', nfirst(X1'''', X2'''')))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




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

ACTIVATE(nfirst(s(X''), cons(Y'', nfirst(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', nfirst(X1'''', X2'''')))
one new Dependency Pair is created:

ACTIVATE(nfirst(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 3
FwdInst
             ...
               →DP Problem 6
Polynomial Ordering
       →DP Problem 2
Nar


Dependency Pairs:

ACTIVATE(nfirst(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))
FIRST(s(X), cons(Y, nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))) -> ACTIVATE(nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

ACTIVATE(nfirst(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  x2  
  POL(FIRST(x1, x2))=  x2  
  POL(s(x1))=  1  
  POL(ACTIVATE(x1))=  x1  
  POL(n__first(x1, x2))=  x1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 3
FwdInst
             ...
               →DP Problem 7
Dependency Graph
       →DP Problem 2
Nar


Dependency Pair:

FIRST(s(X), cons(Y, nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))) -> ACTIVATE(nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Narrowing Transformation


Dependency Pair:

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




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

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))
three new Dependency Pairs are created:

SEL(s(X), cons(Y, nfrom(X''))) -> SEL(X, from(X''))
SEL(s(X), cons(Y, nfirst(X1', X2'))) -> SEL(X, first(X1', X2'))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 8
Narrowing Transformation


Dependency Pairs:

SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, nfirst(X1', X2'))) -> SEL(X, first(X1', X2'))
SEL(s(X), cons(Y, nfrom(X''))) -> SEL(X, from(X''))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




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

SEL(s(X), cons(Y, nfrom(X''))) -> SEL(X, from(X''))
two new Dependency Pairs are created:

SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, nfrom(X'''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 8
Nar
             ...
               →DP Problem 9
Narrowing Transformation


Dependency Pairs:

SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))
SEL(s(X), cons(Y, nfirst(X1', X2'))) -> SEL(X, first(X1', X2'))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




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

SEL(s(X), cons(Y, nfirst(X1', X2'))) -> SEL(X, first(X1', X2'))
three new Dependency Pairs are created:

SEL(s(X), cons(Y, nfirst(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, nfirst(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', nfirst(X'', activate(Z'))))
SEL(s(X), cons(Y, nfirst(X1'', X2''))) -> SEL(X, nfirst(X1'', X2''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 8
Nar
             ...
               →DP Problem 10
Forward Instantiation Transformation


Dependency Pairs:

SEL(s(X), cons(Y, nfirst(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', nfirst(X'', activate(Z'))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




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

SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
three new Dependency Pairs are created:

SEL(s(s(X'')), cons(Y, cons(Y'', Z'''))) -> SEL(s(X''), cons(Y'', Z'''))
SEL(s(s(X'')), cons(Y, cons(Y'', nfrom(X''''')))) -> SEL(s(X''), cons(Y'', nfrom(X''''')))
SEL(s(s(X'')), cons(Y, cons(Y'', nfirst(s(X''''), cons(Y'''', Z'''))))) -> SEL(s(X''), cons(Y'', nfirst(s(X''''), cons(Y'''', Z'''))))

The transformation is resulting in three new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 8
Nar
             ...
               →DP Problem 11
Polynomial Ordering


Dependency Pair:

SEL(s(X), cons(Y, nfirst(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', nfirst(X'', activate(Z'))))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, nfirst(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', nfirst(X'', activate(Z'))))


Additionally, the following usable rules for innermost can be oriented:

activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__from(x1))=  0  
  POL(from(x1))=  0  
  POL(activate(x1))=  x1  
  POL(first(x1, x2))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  0  
  POL(SEL(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  
  POL(n__first(x1, x2))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 8
Nar
             ...
               →DP Problem 14
Dependency Graph


Dependency Pair:


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 8
Nar
             ...
               →DP Problem 12
Polynomial Ordering


Dependency Pair:

SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))


There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 8
Nar
             ...
               →DP Problem 13
Polynomial Ordering


Dependency Pair:

SEL(s(s(X'')), cons(Y, cons(Y'', Z'''))) -> SEL(s(X''), cons(Y'', Z'''))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

SEL(s(s(X'')), cons(Y, cons(Y'', Z'''))) -> SEL(s(X''), cons(Y'', Z'''))


There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem.


Innermost Termination of R successfully shown.
Duration:
0:01 minutes