Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

POL(DOUBLE(x₁)) = 1 + 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

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

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

Dependency Pair:

HALF(s(s(x))) -> HALF(x)

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

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

HALF(s(s(x))) -> HALF(x)

one new Dependency Pair is created:

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

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

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

one new Dependency Pair is created:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

The following dependency pair can be strictly oriented:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

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

POL(HALF(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

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

HALF(x₁) -> HALF(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

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

Dependency Pair:

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

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

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 10

             ↳Forward Instantiation Transformation

Dependency Pair:

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

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

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 10

             ↳FwdInst

...

               →DP Problem 11

                 ↳Argument Filtering and Ordering

Dependency Pair:

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

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

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: Polynomial ordering with Polynomial interpretation:

POL(-'(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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 10

             ↳FwdInst

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pair:

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes

POL(-'(x₁, x₂))	= 1 + x₁ + x₂
POL(s(x₁))	= 1 + x₁