Termination Proof using AProVE

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

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
DOUBLE(s(x)) -> DOUBLE(x)
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> MINUS(x, y)
PLUS(s(x), y) -> DOUBLE(y)
PLUS(s(plus(x, y)), z) -> PLUS(plus(x, y), z)

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

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

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:

DOUBLE(s(x)) -> DOUBLE(x)

Rules:

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

The following dependency pair can be strictly oriented:

DOUBLE(s(x)) -> DOUBLE(x)

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(DOUBLE(x₁)) = 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)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

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

Rules:

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

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

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

five new Dependency Pairs are created:

PLUS(s(plus(0, y'')), z) -> PLUS(y'', z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(plus(x'', s(y'')), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)

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

             ↳Polynomial Ordering

Dependency Pairs:

PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(plus(x'', s(y'')), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(plus(0, y'')), z) -> PLUS(y'', z)
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))

Rules:

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

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(plus(x₁, x₂)) = 1 + x₂

POL(PLUS(x₁, x₂)) = 1 + x₁

POL(0) = 0

POL(minus(x₁, x₂)) = x₁

POL(s(x₁)) = x₁

POL(double(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Polo

...

               →DP Problem 7

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

PLUS(s(plus(s(plus(x'', y'')), y0)), z) -> PLUS(s(plus(plus(x'', y''), y0)), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(minus(x'', y''), double(y''))), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(plus(x'', s(y'')), z)
PLUS(s(plus(s(x''), y'')), z) -> PLUS(s(plus(x'', y'')), z)
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))

Rules:

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

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

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