Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(+(x, y)) -> +'(minus(y), minus(x))
MINUS(+(x, y)) -> MINUS(y)
MINUS(+(x, y)) -> MINUS(x)
*'(x, +(y, z)) -> +'(*(x, y), *(x, z))
*'(x, +(y, z)) -> *'(x, y)
*'(x, +(y, z)) -> *'(x, z)
*'(x, minus(y)) -> MINUS(*(x, y))
*'(x, minus(y)) -> *'(x, y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

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

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

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

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

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

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

The following usable rules for innermost can be oriented:

+(x, 0) -> x
+(minus(x), x) -> 0

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(MINUS(x₁)) = 1 + x₁

POL(minus(x₁)) = x₁

POL(+(x₁, x₂)) = 1 + x₁ + x₂

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

MINUS(x₁) -> MINUS(x₁)
+(x₁, x₂) -> +(x₁, x₂)
minus(x₁) -> minus(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency Pairs:

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

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

The following usable rules for innermost can be oriented:

+(x, 0) -> x
+(minus(x), x) -> 0

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(*'(x₁, x₂)) = 1 + x₁ + x₂

POL(minus(x₁)) = x₁

POL(+(x₁, x₂)) = 1 + x₁ + x₂

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

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(0)	= 0
POL(MINUS(x₁))	= 1 + x₁
POL(minus(x₁))	= x₁
POL(+(x₁, x₂))	= 1 + x₁ + x₂

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