Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(f(b, c, x)) -> F(x, x, x)
ACTIVE(f(x, y, z)) -> F(x, y, active(z))
ACTIVE(f(x, y, z)) -> ACTIVE(z)
F(x, y, mark(z)) -> F(x, y, z)
F(ok(x), ok(y), ok(z)) -> F(x, y, z)
PROPER(f(x, y, z)) -> F(proper(x), proper(y), proper(z))
PROPER(f(x, y, z)) -> PROPER(x)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(z)
TOP(mark(x)) -> TOP(proper(x))
TOP(mark(x)) -> PROPER(x)
TOP(ok(x)) -> TOP(active(x))
TOP(ok(x)) -> ACTIVE(x)

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pairs:

F(ok(x), ok(y), ok(z)) -> F(x, y, z)
F(x, y, mark(z)) -> F(x, y, z)

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(mark(x₁)) = 0

POL(ok(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 5

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(mark(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃)) = x₃

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 5

             ↳Polo

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

ACTIVE(f(x, y, z)) -> ACTIVE(z)

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

The following dependency pair can be strictly oriented:

ACTIVE(f(x, y, z)) -> ACTIVE(z)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(ACTIVE(x₁)) = x₁

POL(f(x₁, x₂, x₃)) = 1 + x₃

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

Dependency Pairs:

PROPER(f(x, y, z)) -> PROPER(z)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(x)

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

The following dependency pairs can be strictly oriented:

PROPER(f(x, y, z)) -> PROPER(z)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(x)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(PROPER(x₁)) = x₁

POL(f(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Narrowing Transformation

Dependency Pairs:

TOP(ok(x)) -> TOP(active(x))
TOP(mark(x)) -> TOP(proper(x))

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

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

TOP(mark(x)) -> TOP(proper(x))

four new Dependency Pairs are created:

TOP(mark(b)) -> TOP(ok(b))
TOP(mark(c)) -> TOP(ok(c))
TOP(mark(d)) -> TOP(ok(d))
TOP(mark(f(x'', y', z'))) -> TOP(f(proper(x''), proper(y'), proper(z')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 9

             ↳Narrowing Transformation

Dependency Pairs:

TOP(mark(f(x'', y', z'))) -> TOP(f(proper(x''), proper(y'), proper(z')))
TOP(mark(d)) -> TOP(ok(d))
TOP(mark(c)) -> TOP(ok(c))
TOP(mark(b)) -> TOP(ok(b))
TOP(ok(x)) -> TOP(active(x))

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

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

TOP(ok(x)) -> TOP(active(x))

four new Dependency Pairs are created:

TOP(ok(f(b, c, x''))) -> TOP(mark(f(x'', x'', x'')))
TOP(ok(f(x'', y', z'))) -> TOP(f(x'', y', active(z')))
TOP(ok(d)) -> TOP(m(b))
TOP(ok(d)) -> TOP(mark(c))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 9

             ↳Nar

...

               →DP Problem 10

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

TOP(ok(f(x'', y', z'))) -> TOP(f(x'', y', active(z')))
TOP(ok(f(b, c, x''))) -> TOP(mark(f(x'', x'', x'')))
TOP(mark(f(x'', y', z'))) -> TOP(f(proper(x''), proper(y'), proper(z')))

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

Termination of R could not be shown.
Duration:
0:00 minutes

POL(mark(x₁))	= 0
POL(ok(x₁))	= 1 + x₁
POL(F(x₁, x₂, x₃))	= x₁

POL(ACTIVE(x₁))	= x₁
POL(f(x₁, x₂, x₃))	= 1 + x₃

POL(PROPER(x₁))	= x₁
POL(f(x₁, x₂, x₃))	= 1 + x₁ + x₂ + x₃