Term Rewriting System R:
[y, x, ys, xs]
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(s(x), y) -> +'(x, y)
++'(:(x, xs), ys) -> ++'(xs, ys)
SUM(:(x, :(y, xs))) -> SUM(:(+(x, y), xs))
SUM(:(x, :(y, xs))) -> +'(x, y)
SUM(++(xs, :(x, :(y, ys)))) -> SUM(++(xs, sum(:(x, :(y, ys)))))
SUM(++(xs, :(x, :(y, ys)))) -> ++'(xs, sum(:(x, :(y, ys))))
SUM(++(xs, :(x, :(y, ys)))) -> SUM(:(x, :(y, ys)))
-'(s(x), s(y)) -> -'(x, y)
QUOT(s(x), s(y)) -> QUOT(-(x, y), s(y))
QUOT(s(x), s(y)) -> -'(x, y)
LENGTH(:(x, xs)) -> LENGTH(xs)
AVG(xs) -> QUOT(hd(sum(xs)), length(xs))
AVG(xs) -> HD(sum(xs))
AVG(xs) -> SUM(xs)
AVG(xs) -> LENGTH(xs)

Furthermore, R contains seven SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

+'(s(x), y) -> +'(x, y)


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





The following dependency pair can be strictly oriented:

+'(s(x), y) -> +'(x, y)


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 8
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

++'(:(x, xs), ys) -> ++'(xs, ys)


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





The following dependency pair can be strictly oriented:

++'(:(x, xs), ys) -> ++'(xs, ys)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(:(x1, x2))=  1 + x2  
  POL(++'(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 9
Dependency Graph
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polynomial Ordering
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





The following dependency pair can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(-'(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 10
Dependency Graph
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polynomial Ordering
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

LENGTH(:(x, xs)) -> LENGTH(xs)


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





The following dependency pair can be strictly oriented:

LENGTH(:(x, xs)) -> LENGTH(xs)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(:(x1, x2))=  1 + x2  
  POL(LENGTH(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
           →DP Problem 11
Dependency Graph
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polynomial Ordering
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

SUM(:(x, :(y, xs))) -> SUM(:(+(x, y), xs))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





The following dependency pair can be strictly oriented:

SUM(:(x, :(y, xs))) -> SUM(:(+(x, y), xs))


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
           →DP Problem 12
Dependency Graph
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polynomial Ordering
       →DP Problem 7
Polo


Dependency Pair:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

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


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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
           →DP Problem 13
Dependency Graph
       →DP Problem 7
Polo


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polynomial Ordering


Dependency Pair:

SUM(++(xs, :(x, :(y, ys)))) -> SUM(++(xs, sum(:(x, :(y, ys)))))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





The following dependency pair can be strictly oriented:

SUM(++(xs, :(x, :(y, ys)))) -> SUM(++(xs, sum(:(x, :(y, ys)))))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))


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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo
           →DP Problem 14
Dependency Graph


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
++(nil, ys) -> ys
++(:(x, xs), ys) -> :(x, ++(xs, ys))
sum(:(x, nil)) -> :(x, nil)
sum(:(x, :(y, xs))) -> sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) -> sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(-(x, y), s(y)))
length(nil) -> 0
length(:(x, xs)) -> s(length(xs))
hd(:(x, xs)) -> x
avg(xs) -> quot(hd(sum(xs)), length(xs))





Using the Dependency Graph resulted in no new DP problems.

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