Termination Proof using AProVE

Term Rewriting System R:
[x, y]
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
MIN(s(x), s(y)) -> MIN(x, y)
TWICE(s(x)) -> TWICE(x)
F(s(x), s(y)) -> F(-(y, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) -> -'(y, min(x, y))
F(s(x), s(y)) -> MIN(x, y)
F(s(x), s(y)) -> TWICE(min(x, y))
F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) -> -'(x, min(x, y))

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

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: Polynomial ordering with Polynomial interpretation:

POL(-'(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MIN(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

Dependency Pair:

TWICE(s(x)) -> TWICE(x)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

TWICE(s(x)) -> TWICE(x)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(TWICE(x₁)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Narrowing Transformation

Dependency Pairs:

F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) -> F(-(y, min(x, y)), s(twice(min(x, y))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(x), s(y)) -> F(-(y, min(x, y)), s(twice(min(x, y))))

six new Dependency Pairs are created:

F(s(x''), s(0)) -> F(-(0, 0), s(twice(min(x'', 0))))
F(s(0), s(y'')) -> F(-(y'', 0), s(twice(min(0, y''))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(twice(0)))
F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(twice(0)))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(twice(0)))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(twice(0)))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(0), s(y'')) -> F(-(y'', 0), s(twice(min(0, y''))))
F(s(x''), s(0)) -> F(-(0, 0), s(twice(min(x'', 0))))
F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(x''), s(0)) -> F(-(0, 0), s(twice(min(x'', 0))))

one new Dependency Pair is created:

F(s(x''), s(0)) -> F(0, s(twice(min(x'', 0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 9

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(twice(0)))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(twice(0)))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(0), s(y'')) -> F(-(y'', 0), s(twice(min(0, y''))))
F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(0), s(y'')) -> F(-(y'', 0), s(twice(min(0, y''))))

one new Dependency Pair is created:

F(s(0), s(y'')) -> F(y'', s(twice(min(0, y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 10

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0), s(y'')) -> F(y'', s(twice(min(0, y''))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(twice(0)))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(twice(0)))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(twice(0)))

one new Dependency Pair is created:

F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 11

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(0))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(twice(0)))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(0), s(y'')) -> F(y'', s(twice(min(0, y''))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(twice(0)))

one new Dependency Pair is created:

F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 12

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(0))
F(s(0), s(y'')) -> F(y'', s(twice(min(0, y''))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(0))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))

six new Dependency Pairs are created:

F(s(x''), s(0)) -> F(-(x'', 0), s(twice(min(x'', 0))))
F(s(0), s(y'')) -> F(-(0, 0), s(twice(min(0, y''))))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(x''), s(0)) -> F(-(x'', min(x'', 0)), s(twice(0)))
F(s(0), s(y'')) -> F(-(0, min(0, y'')), s(twice(0)))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 13

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x'')), s(s(y''))) -> F(-(s(x''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(0), s(y'')) -> F(-(0, min(0, y'')), s(twice(0)))
F(s(x''), s(0)) -> F(-(x'', min(x'', 0)), s(twice(0)))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(0), s(y'')) -> F(-(0, 0), s(twice(min(0, y''))))
F(s(x''), s(0)) -> F(-(x'', 0), s(twice(min(x'', 0))))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(0))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(0), s(y'')) -> F(y'', s(twice(min(0, y''))))
F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(0))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(x''), s(0)) -> F(-(x'', 0), s(twice(min(x'', 0))))

one new Dependency Pair is created:

F(s(x''), s(0)) -> F(x'', s(twice(min(x'', 0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 14

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(x''), s(0)) -> F(x'', s(twice(min(x'', 0))))
F(s(0), s(y'')) -> F(-(0, min(0, y'')), s(twice(0)))
F(s(x''), s(0)) -> F(-(x'', min(x'', 0)), s(twice(0)))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(0), s(y'')) -> F(-(0, 0), s(twice(min(0, y''))))
F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(0))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(0))
F(s(0), s(y'')) -> F(y'', s(twice(min(0, y''))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(0), s(y'')) -> F(-(0, 0), s(twice(min(0, y''))))

one new Dependency Pair is created:

F(s(0), s(y'')) -> F(0, s(twice(min(0, y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 15

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x'')), s(s(y''))) -> F(-(s(x''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(0), s(y'')) -> F(-(0, min(0, y'')), s(twice(0)))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(x''), s(0)) -> F(-(x'', min(x'', 0)), s(twice(0)))
F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(0))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(0))
F(s(0), s(y'')) -> F(y'', s(twice(min(0, y''))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(x''), s(0)) -> F(x'', s(twice(min(x'', 0))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(x''), s(0)) -> F(-(x'', min(x'', 0)), s(twice(0)))

one new Dependency Pair is created:

F(s(x''), s(0)) -> F(-(x'', min(x'', 0)), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 16

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(x''), s(0)) -> F(-(x'', min(x'', 0)), s(0))
F(s(x''), s(0)) -> F(x'', s(twice(min(x'', 0))))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(0), s(y'')) -> F(-(0, min(0, y'')), s(twice(0)))
F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(0))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(0))
F(s(0), s(y'')) -> F(y'', s(twice(min(0, y''))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

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

F(s(0), s(y'')) -> F(-(0, min(0, y'')), s(twice(0)))

one new Dependency Pair is created:

F(s(0), s(y'')) -> F(-(0, min(0, y'')), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 17

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

F(s(0), s(y'')) -> F(-(0, min(0, y'')), s(0))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(s(x'')), s(s(y''))) -> F(-(s(x''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(x''), s(0)) -> F(x'', s(twice(min(x'', 0))))
F(s(0), s(y'')) -> F(-(y'', min(0, y'')), s(0))
F(s(x''), s(0)) -> F(-(0, min(x'', 0)), s(0))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), min(s(x''), s(y''))), s(twice(s(min(x'', y'')))))
F(s(s(x'')), s(s(y''))) -> F(-(s(y''), s(min(x'', y''))), s(twice(min(s(x''), s(y'')))))
F(s(0), s(y'')) -> F(y'', s(twice(min(0, y''))))
F(s(x''), s(0)) -> F(-(x'', min(x'', 0)), s(0))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:04 minutes