Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(app(., app(app(., x), y)), z) -> APP(., y)
APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
APP(i, app(app(., x), y)) -> APP(., app(i, y))
APP(i, app(app(., x), y)) -> APP(i, y)
APP(i, app(app(., x), y)) -> APP(i, x)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

APP(i, app(app(., x), y)) -> APP(i, y)
APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Strategy:

innermost

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

APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))

three new Dependency Pairs are created:

APP(i, app(app(., x), app(i, x''))) -> APP(app(., x''), app(i, x))
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x''))), app(i, x))
APP(i, app(app(., app(i, x'')), y)) -> APP(app(., app(i, y)), x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

APP(i, app(app(., app(i, x'')), y)) -> APP(app(., app(i, y)), x'')
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x''))), app(i, x))
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(i, app(app(., x), app(i, x''))) -> APP(app(., x''), app(i, x))
APP(i, app(app(., x), y)) -> APP(i, y)

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Strategy:

innermost

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

APP(i, app(app(., x), app(i, x''))) -> APP(app(., x''), app(i, x))

one new Dependency Pair is created:

APP(i, app(app(., app(i, x''')), app(i, x''))) -> APP(app(., x''), x''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

APP(i, app(app(., app(i, x''')), app(i, x''))) -> APP(app(., x''), x''')
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x''))), app(i, x))
APP(i, app(app(., x), y)) -> APP(i, y)
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(i, app(app(., app(i, x'')), y)) -> APP(app(., app(i, y)), x'')

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Strategy:

innermost

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

APP(i, app(app(., app(i, x'')), y)) -> APP(app(., app(i, y)), x'')

two new Dependency Pairs are created:

APP(i, app(app(., app(i, x'')), app(i, x'))) -> APP(app(., x'), x'')
APP(i, app(app(., app(i, x'')), app(app(., x'), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x'))), x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Narrowing Transformation

Dependency Pairs:

APP(i, app(app(., app(i, x'')), app(app(., x'), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x'))), x'')
APP(i, app(app(., app(i, x'')), app(i, x'))) -> APP(app(., x'), x'')
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x''))), app(i, x))
APP(i, app(app(., x), y)) -> APP(i, y)
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(i, app(app(., app(i, x''')), app(i, x''))) -> APP(app(., x''), x''')

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Strategy:

innermost

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

APP(i, app(app(., app(i, x''')), app(i, x''))) -> APP(app(., x''), x''')

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Narrowing Transformation

Dependency Pairs:

APP(i, app(app(., app(i, x'')), app(i, x'))) -> APP(app(., x'), x'')
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x''))), app(i, x))
APP(i, app(app(., x), y)) -> APP(i, y)
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(i, app(app(., app(i, x'')), app(app(., x'), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x'))), x'')

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Strategy:

innermost

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

APP(i, app(app(., app(i, x'')), app(i, x'))) -> APP(app(., x'), x'')

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

APP(i, app(app(., app(i, x'')), app(app(., x'), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x'))), x'')
APP(i, app(app(., x), y)) -> APP(i, y)
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x''))), app(i, x))

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Strategy:

innermost

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

APP(i, app(app(., x), y)) -> APP(i, y)

three new Dependency Pairs are created:

APP(i, app(app(., x), app(app(., x''), y''))) -> APP(i, app(app(., x''), y''))
APP(i, app(app(., x), app(app(., x''), app(app(., x''''), y'''')))) -> APP(i, app(app(., x''), app(app(., x''''), y'''')))
APP(i, app(app(., x), app(app(., app(i, x'''')), app(app(., x'0'), y'''')))) -> APP(i, app(app(., app(i, x'''')), app(app(., x'0'), y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

APP(i, app(app(., x), app(app(., app(i, x'''')), app(app(., x'0'), y'''')))) -> APP(i, app(app(., app(i, x'''')), app(app(., x'0'), y'''')))
APP(i, app(app(., x), app(app(., x''), app(app(., x''''), y'''')))) -> APP(i, app(app(., x''), app(app(., x''''), y'''')))
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(i, app(app(., x''), y''))
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x''))), app(i, x))
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(i, app(app(., app(i, x'')), app(app(., x'), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x'))), x'')

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

APP(i, app(app(., x), app(app(., app(i, x'''')), app(app(., x'0'), y'''')))) -> APP(i, app(app(., app(i, x'''')), app(app(., x'0'), y'''')))
APP(i, app(app(., x), app(app(., x''), app(app(., x''''), y'''')))) -> APP(i, app(app(., x''), app(app(., x''''), y'''')))
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(i, app(app(., x''), y''))
APP(i, app(app(., x), app(app(., x''), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x''))), app(i, x))
APP(i, app(app(., app(i, x'')), app(app(., x'), y''))) -> APP(app(., app(app(., app(i, y'')), app(i, x'))), x'')

Additionally, the following usable rules for innermost can be oriented:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(i) = 0

POL(1) = 0

POL(.) = 1

POL(app(x₁, x₂)) = x₁ + x₂

POL(APP(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Polynomial Ordering

Dependency Pair:

APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)

Additionally, the following usable rules for innermost can be oriented:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(i) = 0

POL(1) = 0

POL(.) = 1

POL(app(x₁, x₂)) = x₁ + x₂

POL(APP(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 9

                 ↳Dependency Graph

Dependency Pair:

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(i)	= 0
POL(1)	= 0
POL(.)	= 1
POL(app(x₁, x₂))	= x₁ + x₂
POL(APP(x₁, x₂))	= 1 + x₂