Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, s(0)))
PLUS(plus(x, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), plus(x, s(0)))

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(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(MINUS(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

The following dependency pair can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = x₁

POL(0) = 1

POL(minus(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 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pairs:

PLUS(plus(x, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), plus(x, s(0)))
PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

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

PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, s(0)))

two new Dependency Pairs are created:

PLUS(minus(x, s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x, s(0)))
PLUS(minus(s(x''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x'', 0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

PLUS(minus(s(x''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(plus(x, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), plus(x, s(0)))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

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

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

four new Dependency Pairs are created:

PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(0, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(0))
PLUS(plus(s(x''), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(plus(x'', s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

PLUS(plus(s(x''), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x'', 0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

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

PLUS(minus(x, s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x, s(0)))

two new Dependency Pairs are created:

PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(plus(0, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(minus(s(x''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x'', 0))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(plus(s(x''), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(plus(x'', s(0))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

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

PLUS(minus(s(x''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x'', 0))

two new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Narrowing Transformation

Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')
PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

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

PLUS(plus(0, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(0))

two new Dependency Pairs are created:

PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳Narrowing Transformation

Dependency Pairs:

PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(plus(x'', s(0))))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

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

PLUS(plus(s(x''), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(plus(x'', s(0))))

four new Dependency Pairs are created:

PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(0), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(0)))
PLUS(plus(s(s(x')), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(plus(x', s(0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳Polynomial Ordering

Dependency Pairs:

PLUS(plus(s(s(x')), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')
PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

The following dependency pair can be strictly oriented:

PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(plus(x₁, x₂)) = x₁ + x₂

POL(PLUS(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(minus(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 12

                 ↳Polynomial Ordering

Dependency Pairs:

PLUS(plus(s(s(x')), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

The following dependency pairs can be strictly oriented:

PLUS(plus(s(0), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(0)))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(plus(x₁, x₂)) = 1 + x₁ + x₂

POL(PLUS(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(minus(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 13

                 ↳Polynomial Ordering

Dependency Pairs:

PLUS(plus(s(s(x')), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(plus(x', s(0)))))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

The following dependency pairs can be strictly oriented:

PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(plus(x₁, x₂)) = x₁ + x₂

POL(PLUS(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(minus(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 14

                 ↳Dependency Graph

Dependency Pairs:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes

POL(QUOT(x₁, x₂))	= x₁
POL(0)	= 1
POL(minus(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁

POL(plus(x₁, x₂))	= x₁ + x₂
POL(PLUS(x₁, x₂))	= 1 + x₁ + x₂
POL(0)	= 0
POL(minus(x₁, x₂))	= 1 + x₁
POL(s(x₁))	= x₁