Term Rewriting System R:
[Z, X, Y]
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

FST(s(X), cons(Y, Z)) -> FST(X, Z)
FROM(X) -> FROM(s(X))
ADD(s(X), Y) -> ADD(X, Y)
LEN(cons(X, Z)) -> LEN(Z)

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:

FST(s(X), cons(Y, Z)) -> FST(X, Z)


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))


Strategy:

innermost




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

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

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

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:

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


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))


Strategy:

innermost




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

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

FST(s(s(s(X''''))), cons(Y, cons(Y''0, cons(Y'''', Z'''')))) -> FST(s(s(X'''')), cons(Y''0, cons(Y'''', Z'''')))

The transformation is resulting in one new DP problem:



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


Dependency Pair:

FST(s(s(s(X''''))), cons(Y, cons(Y''0, cons(Y'''', Z'''')))) -> FST(s(s(X'''')), cons(Y''0, cons(Y'''', Z'''')))


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

FST(s(s(s(X''''))), cons(Y, cons(Y''0, cons(Y'''', Z'''')))) -> FST(s(s(X'''')), cons(Y''0, cons(Y'''', Z'''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(FST(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 5
FwdInst
             ...
               →DP Problem 7
Dependency Graph
       →DP Problem 2
Inst
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pair:


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))


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:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))


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:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))


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:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))


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:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))


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:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))


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