Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2, X3]
active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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(X, g(X), Y)) -> F(Y, Y, Y)
ACTIVE(g(X)) -> G(active(X))
ACTIVE(g(X)) -> ACTIVE(X)
G(mark(X)) -> G(X)
G(ok(X)) -> G(X)
PROPER(f(X1, X2, X3)) -> F(proper(X1), proper(X2), proper(X3))
PROPER(f(X1, X2, X3)) -> PROPER(X1)
PROPER(f(X1, X2, X3)) -> PROPER(X2)
PROPER(f(X1, X2, X3)) -> PROPER(X3)
PROPER(g(X)) -> G(proper(X))
PROPER(g(X)) -> PROPER(X)
F(ok(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)
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 five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

F(ok(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

F(ok(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)

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(ok(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

G(ok(X)) -> G(X)
G(mark(X)) -> G(X)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

G(ok(X)) -> G(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(G(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(ok(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

G(mark(X)) -> G(X)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

G(mark(X)) -> G(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(G(x₁)) = x₁

POL(mark(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Polo

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

ACTIVE(g(X)) -> ACTIVE(X)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(g(X)) -> ACTIVE(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(ACTIVE(x₁)) = x₁

POL(g(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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

         ↳Polynomial Ordering

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

PROPER(g(X)) -> PROPER(X)
PROPER(f(X1, X2, X3)) -> PROPER(X3)
PROPER(f(X1, X2, X3)) -> PROPER(X2)
PROPER(f(X1, X2, X3)) -> PROPER(X1)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

PROPER(g(X)) -> 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(g(x₁)) = 1 + x₁

POL(PROPER(x₁)) = x₁

POL(f(x₁, x₂, x₃)) = x₁ + x₂ + x₃

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

           →DP Problem 10

             ↳Polynomial Ordering

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

PROPER(f(X1, X2, X3)) -> PROPER(X3)
PROPER(f(X1, X2, X3)) -> PROPER(X2)
PROPER(f(X1, X2, X3)) -> PROPER(X1)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

PROPER(f(X1, X2, X3)) -> PROPER(X3)
PROPER(f(X1, X2, X3)) -> PROPER(X2)
PROPER(f(X1, X2, X3)) -> PROPER(X1)

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 4

         ↳Polo

           →DP Problem 10

             ↳Polo

...

               →DP Problem 11

                 ↳Dependency Graph

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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

         ↳Polo

       →DP Problem 5

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

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

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

POL(G(x₁))	= x₁
POL(mark(x₁))	= x₁
POL(ok(x₁))	= 1 + x₁

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