Termination Proof using AProVE

Term Rewriting System R:
[x, y]
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)))

Innermost 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)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →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)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and 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)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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.
Used Argument Filtering System:

DOUBLE(x₁) -> DOUBLE(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pairs:

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)))

Strategy:

innermost

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

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

four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Rewriting Transformation

Dependency Pairs:

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

Strategy:

innermost

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

PLUS(s(x''), 0) -> PLUS(x'', double(0))

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 7

                 ↳Rewriting Transformation

Dependency Pairs:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 8

                 ↳Rewriting Transformation

Dependency Pairs:

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

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)))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 9

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:02 minutes

POL(MINUS(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁