Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

NORM(g(x, y)) -> NORM(x)
F(x, g(y, z)) -> F(x, y)
REM(g(x, y), s(z)) -> REM(x, z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

NORM(g(x, y)) -> NORM(x)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

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

NORM(g(x, y)) -> NORM(x)

one new Dependency Pair is created:

NORM(g(g(x'', y''), y)) -> NORM(g(x'', y''))

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:

NORM(g(g(x'', y''), y)) -> NORM(g(x'', y''))

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

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

NORM(g(g(x'', y''), y)) -> NORM(g(x'', y''))

one new Dependency Pair is created:

NORM(g(g(g(x'''', y''''), y''0), y)) -> NORM(g(g(x'''', y''''), y''0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

NORM(g(g(g(x'''', y''''), y''0), y)) -> NORM(g(g(x'''', y''''), y''0))

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

NORM(g(g(g(x'''', y''''), y''0), y)) -> NORM(g(g(x'''', y''''), y''0))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(g(x₁, x₂)) = 1 + x₁

POL(NORM(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳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:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, 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:

F(x, g(y, z)) -> F(x, y)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

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

F(x, g(y, z)) -> F(x, y)

one new Dependency Pair is created:

F(x'', g(g(y'', z''), z)) -> F(x'', g(y'', z''))

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:

F(x'', g(g(y'', z''), z)) -> F(x'', g(y'', z''))

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

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

F(x'', g(g(y'', z''), z)) -> F(x'', g(y'', z''))

one new Dependency Pair is created:

F(x'''', g(g(g(y'''', z''''), z''0), z)) -> F(x'''', g(g(y'''', z''''), z''0))

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

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

F(x'''', g(g(g(y'''', z''''), z''0), z)) -> F(x'''', g(g(y'''', z''''), z''0))

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(x'''', g(g(g(y'''', z''''), z''0), z)) -> F(x'''', g(g(y'''', z''''), z''0))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(g(x₁, x₂)) = 1 + x₁

POL(F(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳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:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, 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:

REM(g(x, y), s(z)) -> REM(x, z)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

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

REM(g(x, y), s(z)) -> REM(x, z)

one new Dependency Pair is created:

REM(g(g(x'', y''), y), s(s(z''))) -> REM(g(x'', y''), s(z''))

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:

REM(g(g(x'', y''), y), s(s(z''))) -> REM(g(x'', y''), s(z''))

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

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

REM(g(g(x'', y''), y), s(s(z''))) -> REM(g(x'', y''), s(z''))

one new Dependency Pair is created:

REM(g(g(g(x'''', y''''), y''0), y), s(s(s(z'''')))) -> REM(g(g(x'''', y''''), y''0), s(s(z'''')))

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

                 ↳Polynomial Ordering

Dependency Pair:

REM(g(g(g(x'''', y''''), y''0), y), s(s(s(z'''')))) -> REM(g(g(x'''', y''''), y''0), s(s(z'''')))

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

REM(g(g(g(x'''', y''''), y''0), y), s(s(s(z'''')))) -> REM(g(g(x'''', y''''), y''0), s(s(z'''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(g(x₁, x₂)) = 0

POL(REM(x₁, x₂)) = x₂

POL(s(x₁)) = 1 + x₁

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 12

                 ↳Dependency Graph

Dependency Pair:

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(g(x₁, x₂))	= 1 + x₁
POL(NORM(x₁))	= 1 + x₁

POL(g(x₁, x₂))	= 1 + x₁
POL(F(x₁, x₂))	= 1 + x₂

POL(g(x₁, x₂))	= 0
POL(REM(x₁, x₂))	= x₂
POL(s(x₁))	= 1 + x₁