Term Rewriting System R:
[k, l, x, y]
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(cons(x, l), k) -> APP(l, k)
SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))
SUM(cons(x, cons(y, l))) -> PLUS(x, y)
SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))
SUM(app(l, cons(x, cons(y, k)))) -> APP(l, sum(cons(x, cons(y, k))))
SUM(app(l, cons(x, cons(y, k)))) -> SUM(cons(x, cons(y, k)))
SUM(plus(cons(0, x), cons(y, l))) -> PRED(sum(cons(s(x), cons(y, l))))
SUM(plus(cons(0, x), cons(y, l))) -> SUM(cons(s(x), cons(y, l)))
PLUS(s(x), y) -> PLUS(x, y)

Furthermore, R contains four SCCs.


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


Dependency Pair:

APP(cons(x, l), k) -> APP(l, k)


Rules:


app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)





The following dependency pair can be strictly oriented:

APP(cons(x, l), k) -> APP(l, k)


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(APP(x1, x2))=  x1 + x2  

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
AFS
           →DP Problem 5
Dependency Graph
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS


Dependency Pair:


Rules:


app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

PLUS(s(x), y) -> PLUS(x, y)


Rules:


app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)





The following dependency pair can be strictly oriented:

PLUS(s(x), y) -> PLUS(x, y)


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(PLUS(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  

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


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 6
Dependency Graph
       →DP Problem 3
AFS
       →DP Problem 4
AFS


Dependency Pair:


Rules:


app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))


Rules:


app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)





The following dependency pair can be strictly oriented:

SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))


The following usable rules using the Ce-refinement can be oriented:

plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))


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

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


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
           →DP Problem 7
Dependency Graph
       →DP Problem 4
AFS


Dependency Pair:


Rules:


app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))


Rules:


app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)





The following dependency pair can be strictly oriented:

SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))


The following usable rules using the Ce-refinement can be oriented:

app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(plus(x1, x2))=  x1 + x2  
  POL(SUM(x1))=  1 + x1  
  POL(0)=  0  
  POL(cons(x1))=  1 + x1  
  POL(pred(x1))=  x1  
  POL(nil)=  0  
  POL(sum)=  1  
  POL(s(x1))=  x1  
  POL(app(x1, x2))=  x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
SUM(x1) -> SUM(x1)
app(x1, x2) -> app(x1, x2)
cons(x1, x2) -> cons(x2)
sum(x1) -> sum
pred(x1) -> pred(x1)
plus(x1, x2) -> plus(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
           →DP Problem 8
Dependency Graph


Dependency Pair:


Rules:


app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)





Using the Dependency Graph resulted in no new DP problems.

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