Termination Proof using AProVE

Term Rewriting System R:
[X, XS, X1, X2]
active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
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(zeros) -> CONS(0, zeros)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(tail(X)) -> TAIL(active(X))
ACTIVE(tail(X)) -> ACTIVE(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
TAIL(mark(X)) -> TAIL(X)
TAIL(ok(X)) -> TAIL(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(tail(X)) -> TAIL(proper(X))
PROPER(tail(X)) -> PROPER(X)
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 Pairs:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)

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(CONS(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

CONS(mark(X1), X2) -> CONS(X1, X2)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

CONS(mark(X1), X2) -> CONS(X1, X2)

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(CONS(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
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:

TAIL(ok(X)) -> TAIL(X)
TAIL(mark(X)) -> TAIL(X)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

TAIL(ok(X)) -> TAIL(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(TAIL(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 8

             ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

TAIL(mark(X)) -> TAIL(X)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

TAIL(mark(X)) -> TAIL(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(TAIL(x₁)) = x₁

POL(mark(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 8

             ↳Polo

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
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 Pairs:

ACTIVE(tail(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(tail(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(cons(x₁, x₂)) = x₁

POL(tail(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 10

             ↳Polynomial Ordering

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(cons(X1, X2)) -> ACTIVE(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(ACTIVE(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 10

             ↳Polo

...

               →DP Problem 11

                 ↳Dependency Graph

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
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(tail(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

PROPER(tail(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(PROPER(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁ + x₂

POL(tail(x₁)) = 1 + x₁

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 12

             ↳Polynomial Ordering

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> 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(cons(x₁, x₂)) = 1 + x₁ + x₂

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 12

             ↳Polo

...

               →DP Problem 13

                 ↳Dependency Graph

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
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(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
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(CONS(x₁, x₂))	= x₂

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

POL(ACTIVE(x₁))	= x₁
POL(cons(x₁, x₂))	= x₁
POL(tail(x₁))	= 1 + x₁

POL(PROPER(x₁))	= x₁
POL(cons(x₁, x₂))	= x₁ + x₂
POL(tail(x₁))	= 1 + x₁