Termination Proof using AProVE

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.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 8

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 11

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 12

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 13

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 14

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 15

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 16

                 ↳Dependency Graph

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 17

             ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 17

             ↳FwdInst

...

               →DP Problem 18

                 ↳Argument Filtering and Ordering

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 17

             ↳FwdInst

...

               →DP Problem 19

                 ↳Dependency Graph

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Argument Filtering and Ordering

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

           →DP Problem 20

             ↳Dependency Graph

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Narrowing Transformation

       →DP Problem 7

         ↳Nar

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

           →DP Problem 21

             ↳Forward Instantiation Transformation

           →DP Problem 22

             ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))

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

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

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))

one new Dependency Pair is created:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

           →DP Problem 21

             ↳FwdInst

...

               →DP Problem 23

                 ↳Argument Filtering and Ordering

           →DP Problem 22

             ↳Nar

       →DP Problem 7

         ↳Nar

Dependency Pair:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))

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(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))

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:

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

           →DP Problem 21

             ↳FwdInst

...

               →DP Problem 26

                 ↳Dependency Graph

           →DP Problem 22

             ↳Nar

       →DP Problem 7

         ↳Nar

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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

           →DP Problem 21

             ↳FwdInst

           →DP Problem 22

             ↳Narrowing Transformation

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

           →DP Problem 21

             ↳FwdInst

           →DP Problem 22

             ↳Nar

...

               →DP Problem 24

                 ↳Argument Filtering and Ordering

       →DP Problem 7

         ↳Nar

Dependency Pair:

QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))

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(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))

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:

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

           →DP Problem 21

             ↳FwdInst

           →DP Problem 22

             ↳Nar

...

               →DP Problem 25

                 ↳Argument Filtering and Ordering

       →DP Problem 7

         ↳Nar

Dependency Pair:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Narrowing Transformation

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

           →DP Problem 29

             ↳Narrowing Transformation

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

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

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

four new Dependency Pairs are created:

SUM(++(xs, :(x''', :(y''', nil)))) -> SUM(++(xs, :(+(x''', y'''), nil)))
SUM(++(xs, :(x''', :(y''', :(y', xs''))))) -> SUM(++(xs, sum(:(+(+(x''', y'''), y'), xs''))))
SUM(++(xs, :(0, :(y''', ys')))) -> SUM(++(xs, sum(:(y''', ys'))))
SUM(++(xs, :(s(x'), :(y''', ys')))) -> SUM(++(xs, sum(:(s(+(x', y''')), ys'))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

           →DP Problem 29

             ↳Nar

...

               →DP Problem 30

                 ↳Argument Filtering and Ordering

Dependency Pairs:

SUM(++(xs, :(s(x'), :(y''', ys')))) -> SUM(++(xs, sum(:(s(+(x', y''')), ys'))))
SUM(++(xs, :(0, :(y''', ys')))) -> SUM(++(xs, sum(:(y''', ys'))))
SUM(++(xs, :(x''', :(y''', :(y', xs''))))) -> SUM(++(xs, sum(:(+(+(x''', y'''), y'), xs''))))
SUM(++(xs, :(x''', :(y''', nil)))) -> SUM(++(xs, :(+(x''', y'''), nil)))

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 pairs can be strictly oriented:

SUM(++(xs, :(s(x'), :(y''', ys')))) -> SUM(++(xs, sum(:(s(+(x', y''')), ys'))))
SUM(++(xs, :(0, :(y''', ys')))) -> SUM(++(xs, sum(:(y''', ys'))))
SUM(++(xs, :(x''', :(y''', :(y', xs''))))) -> SUM(++(xs, sum(:(+(+(x''', y'''), y'), xs''))))
SUM(++(xs, :(x''', :(y''', nil)))) -> SUM(++(xs, :(+(x''', y'''), nil)))

The following usable rules for innermost 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)))))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))

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

++ > :

resulting in one new DP problem.
Used Argument Filtering System:

SUM(x₁) -> SUM(x₁)
++(x₁, x₂) -> ++(x₁, x₂)
:(x₁, x₂) -> :(x₁, x₂)
sum(x₁) -> x₁
+(x₁, x₂) -> x₂
s(x₁) -> x₁

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Nar

       →DP Problem 7

         ↳Nar

           →DP Problem 29

             ↳Nar

...

               →DP Problem 31

                 ↳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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:07 minutes