Term Rewriting System R:
[X, Y, M, N]
filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)
FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
SIEVE(cons(0, Y)) -> SIEVE(Y)
SIEVE(cons(s(N), Y)) -> SIEVE(filter(Y, N, N))
SIEVE(cons(s(N), Y)) -> FILTER(Y, N, N)
NATS(N) -> NATS(s(N))
ZPRIMES -> SIEVE(nats(s(s(0))))
ZPRIMES -> NATS(s(s(0)))

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Forward Instantiation Transformation
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pairs:

FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)
two new Dependency Pairs are created:

FILTER(cons(X, cons(X'', Y'')), 0, 0) -> FILTER(cons(X'', Y''), 0, 0)
FILTER(cons(X, cons(X'', Y'')), 0, s(N'')) -> FILTER(cons(X'', Y''), s(N''), s(N''))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
Forward Instantiation Transformation
           →DP Problem 5
FwdInst
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pair:

FILTER(cons(X, cons(X'', Y'')), 0, 0) -> FILTER(cons(X'', Y''), 0, 0)


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

FILTER(cons(X, cons(X'', Y'')), 0, 0) -> FILTER(cons(X'', Y''), 0, 0)
one new Dependency Pair is created:

FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), 0, 0) -> FILTER(cons(X'''', cons(X''''', Y'''')), 0, 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 6
Polynomial Ordering
           →DP Problem 5
FwdInst
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pair:

FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), 0, 0) -> FILTER(cons(X'''', cons(X''''', Y'''')), 0, 0)


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), 0, 0) -> FILTER(cons(X'''', cons(X''''', Y'''')), 0, 0)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(FILTER(x1, x2, x3))=  1 + x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 12
Dependency Graph
           →DP Problem 5
FwdInst
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pair:


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
           →DP Problem 5
Forward Instantiation Transformation
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pairs:

FILTER(cons(X, cons(X'', Y'')), 0, s(N'')) -> FILTER(cons(X'', Y''), s(N''), s(N''))
FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
two new Dependency Pairs are created:

FILTER(cons(X, cons(X'', Y'')), s(s(N'')), M'') -> FILTER(cons(X'', Y''), s(N''), M'')
FILTER(cons(X, cons(X'', cons(X'''', Y''''))), s(0), s(N'''')) -> FILTER(cons(X'', cons(X'''', Y'''')), 0, s(N''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 7
Forward Instantiation Transformation
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pairs:

FILTER(cons(X, cons(X'', cons(X'''', Y''''))), s(0), s(N'''')) -> FILTER(cons(X'', cons(X'''', Y'''')), 0, s(N''''))
FILTER(cons(X, cons(X'', Y'')), s(s(N'')), M'') -> FILTER(cons(X'', Y''), s(N''), M'')
FILTER(cons(X, cons(X'', Y'')), 0, s(N'')) -> FILTER(cons(X'', Y''), s(N''), s(N''))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

FILTER(cons(X, cons(X'', Y'')), 0, s(N'')) -> FILTER(cons(X'', Y''), s(N''), s(N''))
two new Dependency Pairs are created:

FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), 0, s(s(N''''))) -> FILTER(cons(X'''', cons(X''''', Y'''')), s(s(N'''')), s(s(N'''')))
FILTER(cons(X, cons(X'''', cons(X'''0, cons(X'''''', Y'''''')))), 0, s(0)) -> FILTER(cons(X'''', cons(X'''0, cons(X'''''', Y''''''))), s(0), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 8
Forward Instantiation Transformation
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pairs:

FILTER(cons(X, cons(X'''', cons(X'''0, cons(X'''''', Y'''''')))), 0, s(0)) -> FILTER(cons(X'''', cons(X'''0, cons(X'''''', Y''''''))), s(0), s(0))
FILTER(cons(X, cons(X'', Y'')), s(s(N'')), M'') -> FILTER(cons(X'', Y''), s(N''), M'')
FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), 0, s(s(N''''))) -> FILTER(cons(X'''', cons(X''''', Y'''')), s(s(N'''')), s(s(N'''')))
FILTER(cons(X, cons(X'', cons(X'''', Y''''))), s(0), s(N'''')) -> FILTER(cons(X'', cons(X'''', Y'''')), 0, s(N''''))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

FILTER(cons(X, cons(X'', Y'')), s(s(N'')), M'') -> FILTER(cons(X'', Y''), s(N''), M'')
two new Dependency Pairs are created:

FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), s(s(s(N''''))), M'''') -> FILTER(cons(X'''', cons(X''''', Y'''')), s(s(N'''')), M'''')
FILTER(cons(X, cons(X'''', cons(X'''0, cons(X'''''', Y'''''')))), s(s(0)), s(N'''''')) -> FILTER(cons(X'''', cons(X'''0, cons(X'''''', Y''''''))), s(0), s(N''''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 9
Forward Instantiation Transformation
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pairs:

FILTER(cons(X, cons(X'''', cons(X'''0, cons(X'''''', Y'''''')))), s(s(0)), s(N'''''')) -> FILTER(cons(X'''', cons(X'''0, cons(X'''''', Y''''''))), s(0), s(N''''''))
FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), s(s(s(N''''))), M'''') -> FILTER(cons(X'''', cons(X''''', Y'''')), s(s(N'''')), M'''')
FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), 0, s(s(N''''))) -> FILTER(cons(X'''', cons(X''''', Y'''')), s(s(N'''')), s(s(N'''')))
FILTER(cons(X, cons(X'', cons(X'''', Y''''))), s(0), s(N'''')) -> FILTER(cons(X'', cons(X'''', Y'''')), 0, s(N''''))
FILTER(cons(X, cons(X'''', cons(X'''0, cons(X'''''', Y'''''')))), 0, s(0)) -> FILTER(cons(X'''', cons(X'''0, cons(X'''''', Y''''''))), s(0), s(0))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

FILTER(cons(X, cons(X'', cons(X'''', Y''''))), s(0), s(N'''')) -> FILTER(cons(X'', cons(X'''', Y'''')), 0, s(N''''))
two new Dependency Pairs are created:

FILTER(cons(X, cons(X''', cons(X''''', cons(X'''''''', Y'''''')))), s(0), s(s(N''''''))) -> FILTER(cons(X''', cons(X''''', cons(X'''''''', Y''''''))), 0, s(s(N'''''')))
FILTER(cons(X, cons(X''', cons(X'''''', cons(X'''0'', cons(X'''''''', Y''''''''))))), s(0), s(0)) -> FILTER(cons(X''', cons(X'''''', cons(X'''0'', cons(X'''''''', Y'''''''')))), 0, s(0))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 10
Polynomial Ordering
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pairs:

FILTER(cons(X, cons(X'''', cons(X'''0, cons(X'''''', Y'''''')))), 0, s(0)) -> FILTER(cons(X'''', cons(X'''0, cons(X'''''', Y''''''))), s(0), s(0))
FILTER(cons(X, cons(X''', cons(X'''''', cons(X'''0'', cons(X'''''''', Y''''''''))))), s(0), s(0)) -> FILTER(cons(X''', cons(X'''''', cons(X'''0'', cons(X'''''''', Y'''''''')))), 0, s(0))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

FILTER(cons(X, cons(X'''', cons(X'''0, cons(X'''''', Y'''''')))), 0, s(0)) -> FILTER(cons(X'''', cons(X'''0, cons(X'''''', Y''''''))), s(0), s(0))
FILTER(cons(X, cons(X''', cons(X'''''', cons(X'''0'', cons(X'''''''', Y''''''''))))), s(0), s(0)) -> FILTER(cons(X''', cons(X'''''', cons(X'''0'', cons(X'''''''', Y'''''''')))), 0, s(0))


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(FILTER(x1, x2, x3))=  1 + x1 + x3  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x2  
  POL(s(x1))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 11
Polynomial Ordering
       →DP Problem 2
Inst
       →DP Problem 3
Remaining


Dependency Pairs:

FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), s(s(s(N''''))), M'''') -> FILTER(cons(X'''', cons(X''''', Y'''')), s(s(N'''')), M'''')
FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), 0, s(s(N''''))) -> FILTER(cons(X'''', cons(X''''', Y'''')), s(s(N'''')), s(s(N'''')))
FILTER(cons(X, cons(X''', cons(X''''', cons(X'''''''', Y'''''')))), s(0), s(s(N''''''))) -> FILTER(cons(X''', cons(X''''', cons(X'''''''', Y''''''))), 0, s(s(N'''''')))
FILTER(cons(X, cons(X'''', cons(X'''0, cons(X'''''', Y'''''')))), s(s(0)), s(N'''''')) -> FILTER(cons(X'''', cons(X'''0, cons(X'''''', Y''''''))), s(0), s(N''''''))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), s(s(s(N''''))), M'''') -> FILTER(cons(X'''', cons(X''''', Y'''')), s(s(N'''')), M'''')
FILTER(cons(X, cons(X'''', cons(X''''', Y''''))), 0, s(s(N''''))) -> FILTER(cons(X'''', cons(X''''', Y'''')), s(s(N'''')), s(s(N'''')))
FILTER(cons(X, cons(X''', cons(X''''', cons(X'''''''', Y'''''')))), s(0), s(s(N''''''))) -> FILTER(cons(X''', cons(X''''', cons(X'''''''', Y''''''))), 0, s(s(N'''''')))
FILTER(cons(X, cons(X'''', cons(X'''0, cons(X'''''', Y'''''')))), s(s(0)), s(N'''''')) -> FILTER(cons(X'''', cons(X'''0, cons(X'''''', Y''''''))), s(0), s(N''''''))


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(FILTER(x1, x2, x3))=  1 + x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x2  
  POL(s(x1))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pair:

NATS(N) -> NATS(s(N))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

NATS(N) -> NATS(s(N))
one new Dependency Pair is created:

NATS(s(N'')) -> NATS(s(s(N'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Inst
           →DP Problem 15
Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pair:

NATS(s(N'')) -> NATS(s(s(N'')))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

NATS(s(N'')) -> NATS(s(s(N'')))
one new Dependency Pair is created:

NATS(s(s(N''''))) -> NATS(s(s(s(N''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Inst
           →DP Problem 15
Inst
             ...
               →DP Problem 16
Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pair:

NATS(s(s(N''''))) -> NATS(s(s(s(N''''))))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

NATS(s(s(N''''))) -> NATS(s(s(s(N''''))))
one new Dependency Pair is created:

NATS(s(s(s(N'''''')))) -> NATS(s(s(s(s(N'''''')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Inst
           →DP Problem 15
Inst
             ...
               →DP Problem 17
Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pair:

NATS(s(s(s(N'''''')))) -> NATS(s(s(s(s(N'''''')))))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

NATS(s(s(s(N'''''')))) -> NATS(s(s(s(s(N'''''')))))
one new Dependency Pair is created:

NATS(s(s(s(s(N''''''''))))) -> NATS(s(s(s(s(s(N''''''''))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Inst
           →DP Problem 15
Inst
             ...
               →DP Problem 18
Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pair:

NATS(s(s(s(s(N''''''''))))) -> NATS(s(s(s(s(s(N''''''''))))))


Rules:


filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))


Strategy:

innermost




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

NATS(s(s(s(s(N''''''''))))) -> NATS(s(s(s(s(s(N''''''''))))))
one new Dependency Pair is created:

NATS(s(s(s(s(s(N'''''''''')))))) -> NATS(s(s(s(s(s(s(N'''''''''')))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Inst
       →DP Problem 3
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Inst
       →DP Problem 3
Remaining Obligation(s)




The following remains to be proven:

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