Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(0(x), 0(y)) -> 0'(+(x, y))
+'(0(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), j(y)) -> +'(x, y)
+'(j(x), 0(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(+(x, y), j(#))
+'(j(x), j(y)) -> +'(x, y)
+'(1(x), j(y)) -> 0'(+(x, y))
+'(1(x), j(y)) -> +'(x, y)
+'(j(x), 1(y)) -> 0'(+(x, y))
+'(j(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
OPP(0(x)) -> 0'(opp(x))
OPP(0(x)) -> OPP(x)
OPP(1(x)) -> OPP(x)
OPP(j(x)) -> OPP(x)
-'(x, y) -> +'(x, opp(y))
-'(x, y) -> OPP(y)
*'(0(x), y) -> 0'(*(x, y))
*'(0(x), y) -> *'(x, y)
*'(1(x), y) -> +'(0(*(x, y)), y)
*'(1(x), y) -> 0'(*(x, y))
*'(1(x), y) -> *'(x, y)
*'(j(x), y) -> -'(0(*(x, y)), y)
*'(j(x), y) -> 0'(*(x, y))
*'(j(x), y) -> *'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(j(x), 1(y)) -> +'(x, y)
+'(1(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(+(x, y), j(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(j(x), 0(y)) -> +'(x, y)
+'(0(x), j(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

+'(j(x), 1(y)) -> +'(x, y)
+'(1(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(+(x, y), j(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(j(x), 0(y)) -> +'(x, y)
+'(0(x), j(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)

The following usable rules for innermost can be oriented:

+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(#) = 0

POL(0(x₁)) = x₁

POL(1(x₁)) = 1 + x₁

POL(j(x₁)) = 1 + x₁

POL(+(x₁, x₂)) = x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

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

+'(x₁, x₂) -> +'(x₁, x₂)
j(x₁) -> j(x₁)
0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)
+(x₁, x₂) -> +(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(0(x), 0(y)) -> +'(x, y)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(0(x), 0(y)) -> +'(x, y)

The following usable rules for innermost can be oriented:

+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(#) = 0

POL(0(x₁)) = 1 + x₁

POL(1(x₁)) = 1 + x₁

POL(j(x₁)) = 1 + x₁

POL(+(x₁, x₂)) = x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

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

+'(x₁, x₂) -> +'(x₁, x₂)
0(x₁) -> 0(x₁)
+(x₁, x₂) -> +(x₁, x₂)
1(x₁) -> 1(x₁)
j(x₁) -> j(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳AFS

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳AFS

...

               →DP Problem 6

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

+'(+(x, y), z) -> +'(y, z)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(+(x, y), z) -> +'(y, z)

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(+'(x₁, x₂)) = x₁ + x₂

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

+'(x₁, x₂) -> +'(x₁, x₂)
+(x₁, x₂) -> +(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳AFS

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

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

         ↳AFS

Dependency Pairs:

OPP(j(x)) -> OPP(x)
OPP(1(x)) -> OPP(x)
OPP(0(x)) -> OPP(x)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

OPP(j(x)) -> OPP(x)

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

POL(0(x₁)) = x₁

POL(1(x₁)) = x₁

POL(OPP(x₁)) = x₁

POL(j(x₁)) = 1 + x₁

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

OPP(x₁) -> OPP(x₁)
j(x₁) -> j(x₁)
0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 8

             ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

Dependency Pairs:

OPP(1(x)) -> OPP(x)
OPP(0(x)) -> OPP(x)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

OPP(0(x)) -> OPP(x)

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

POL(0(x₁)) = 1 + x₁

POL(1(x₁)) = x₁

POL(OPP(x₁)) = x₁

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

OPP(x₁) -> OPP(x₁)
0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 8

             ↳AFS

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

Dependency Pair:

OPP(1(x)) -> OPP(x)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

OPP(1(x)) -> OPP(x)

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

POL(1(x₁)) = 1 + x₁

POL(OPP(x₁)) = x₁

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

OPP(x₁) -> OPP(x₁)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 8

             ↳AFS

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

Dependency Pairs:

*'(*(x, y), z) -> *'(y, z)
*'(j(x), y) -> *'(x, y)
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

*'(*(x, y), z) -> *'(y, z)

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

POL(0(x₁)) = x₁

POL(*'(x₁, x₂)) = x₁ + x₂

POL(1(x₁)) = x₁

POL(*(x₁, x₂)) = 1 + x₁ + x₂

POL(j(x₁)) = x₁

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

*'(x₁, x₂) -> *'(x₁, x₂)
*(x₁, x₂) -> *(x₁, x₂)
j(x₁) -> j(x₁)
0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 11

             ↳Argument Filtering and Ordering

Dependency Pairs:

*'(j(x), y) -> *'(x, y)
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

*'(j(x), y) -> *'(x, y)

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

POL(0(x₁)) = x₁

POL(*'(x₁, x₂)) = x₁ + x₂

POL(1(x₁)) = x₁

POL(j(x₁)) = 1 + x₁

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

*'(x₁, x₂) -> *'(x₁, x₂)
j(x₁) -> j(x₁)
0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 11

             ↳AFS

...

               →DP Problem 12

                 ↳Argument Filtering and Ordering

Dependency Pairs:

*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

*'(0(x), y) -> *'(x, y)

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

POL(0(x₁)) = 1 + x₁

POL(*'(x₁, x₂)) = x₁ + x₂

POL(1(x₁)) = x₁

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

*'(x₁, x₂) -> *'(x₁, x₂)
0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 11

             ↳AFS

...

               →DP Problem 13

                 ↳Argument Filtering and Ordering

Dependency Pair:

*'(1(x), y) -> *'(x, y)

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

*'(1(x), 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 + x₂

POL(1(x₁)) = 1 + x₁

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

*'(x₁, x₂) -> *'(x₁, x₂)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 11

             ↳AFS

...

               →DP Problem 14

                 ↳Dependency Graph

Dependency Pair:

Rules:

0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes

POL(#)	= 0
POL(0(x₁))	= x₁
POL(1(x₁))	= 1 + x₁
POL(j(x₁))	= 1 + x₁
POL(+(x₁, x₂))	= x₁ + x₂
POL(+'(x₁, x₂))	= 1 + x₁ + x₂

POL(0(x₁))	= x₁
POL(*'(x₁, x₂))	= x₁ + x₂
POL(1(x₁))	= x₁
POL(*(x₁, x₂))	= 1 + x₁ + x₂
POL(j(x₁))	= x₁