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))

Innermost 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
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
AFS
       →DP Problem 6
AFS
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


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


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))


Strategy:

innermost




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
       →DP Problem 5
AFS
       →DP Problem 6
AFS
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


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


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))


Strategy:

innermost




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
       →DP Problem 5
AFS
       →DP Problem 6
AFS
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


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


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))


Strategy:

innermost




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
       →DP Problem 5
AFS
       →DP Problem 6
AFS
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


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


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))


Strategy:

innermost




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
AFS
       →DP Problem 5
Argument Filtering and Ordering
       →DP Problem 6
AFS
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


The following usable rules for innermost can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


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


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))


Strategy:

innermost




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
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Argument Filtering and Ordering
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


The following usable rules for innermost can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


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


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))


Strategy:

innermost




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
AFS
       →DP Problem 5
AFS
       →DP Problem 6
AFS
       →DP Problem 7
Remaining Obligation(s)




The following remains to be proven:
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))


Strategy:

innermost



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