Term Rewriting System R:
[YS, X, XS, Y, L]
app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(cons(X, XS), YS) -> APP(XS, YS)
FROM(X) -> FROM(s(X))
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, nil))
ZWADR(cons(X, XS), cons(Y, YS)) -> ZWADR(XS, YS)
PREFIX(L) -> ZWADR(L, prefix(L))
PREFIX(L) -> PREFIX(L)

Furthermore, R contains four SCCs. 


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


Dependency Pair:

APP(cons(X, XS), YS) -> APP(XS, YS)


Rules:


app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))


Strategy:

innermost




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

APP(cons(X, XS), YS) -> APP(XS, YS)
one new Dependency Pair is created:

APP(cons(X, cons(X'', XS'')), YS'') -> APP(cons(X'', XS''), YS'')

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

APP(cons(X, cons(X'', XS'')), YS'') -> APP(cons(X'', XS''), YS'')


Rules:


app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))


Strategy:

innermost




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

APP(cons(X, cons(X'', XS'')), YS'') -> APP(cons(X'', XS''), YS'')
one new Dependency Pair is created:

APP(cons(X, cons(X'''', cons(X''''', XS''''))), YS'''') -> APP(cons(X'''', cons(X''''', XS'''')), YS'''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 5
             ↳FwdInst
             ...
               →DP Problem 6
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Inst
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

APP(cons(X, cons(X'''', cons(X''''', XS''''))), YS'''') -> APP(cons(X'''', cons(X''''', XS'''')), YS'''')


Rules:


app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

APP(cons(X, cons(X'''', cons(X''''', XS''''))), YS'''') -> APP(cons(X'''', cons(X''''', XS'''')), YS'''')


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> APP(x1, x2)
cons(x1, x2) -> cons(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 5
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Dependency Graph
       →DP Problem 2
         ↳Inst
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

FROM(X) -> FROM(s(X))


Rules:


app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))


Strategy:

innermost




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

FROM(X) -> FROM(s(X))
one new Dependency Pair is created:

FROM(s(X'')) -> FROM(s(s(X'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Inst
           →DP Problem 8
             ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

FROM(s(X'')) -> FROM(s(s(X'')))


Rules:


app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))


Strategy:

innermost




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

FROM(s(X'')) -> FROM(s(s(X'')))
one new Dependency Pair is created:

FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Inst
           →DP Problem 8
             ↳Inst
             ...
               →DP Problem 9
                 ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))


Rules:


app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))


Strategy:

innermost




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

FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))
one new Dependency Pair is created:

FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Inst
           →DP Problem 8
             ↳Inst
             ...
               →DP Problem 10
                 ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))


Rules:


app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))


Strategy:

innermost




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

FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))
one new Dependency Pair is created:

FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Inst
           →DP Problem 8
             ↳Inst
             ...
               →DP Problem 11
                 ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))


Rules:


app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))


Strategy:

innermost




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

FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))
one new Dependency Pair is created:

FROM(s(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(s(s(X'''''''''')))))))

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)
       →DP Problem 4
         ↳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)
       →DP Problem 4
         ↳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)
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:

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