Termination Proof using AProVE

Term Rewriting System R:
[X, X1, X2, X3]
active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(f(a, b, X)) -> F(X, X, X)
ACTIVE(f(X1, X2, X3)) -> F(active(X1), X2, X3)
ACTIVE(f(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(f(X1, X2, X3)) -> F(X1, X2, active(X3))
ACTIVE(f(X1, X2, X3)) -> ACTIVE(X3)
F(mark(X1), X2, X3) -> F(X1, X2, X3)
F(X1, X2, mark(X3)) -> F(X1, X2, X3)
F(ok(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)
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)
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(X1, X2, mark(X3)) -> F(X1, X2, X3)
F(mark(X1), X2, X3) -> F(X1, X2, X3)
F(ok(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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)) = x₂

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 Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 5

             ↳Polo

...

               →DP Problem 6

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

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

Strategy:

innermost

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 Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost 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 + x₁ + x₃

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

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

Strategy:

innermost

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

Rules:

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

Strategy:

innermost

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 for innermost 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 9

             ↳Dependency Graph

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

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

Strategy:

innermost

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

Strategy:

innermost

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(f(X1', X2', X3'))) -> TOP(f(proper(X1'), proper(X2'), proper(X3')))
TOP(mark(a)) -> TOP(ok(a))
TOP(mark(b)) -> TOP(ok(b))
TOP(mark(c)) -> TOP(ok(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 10

             ↳Narrowing Transformation

Dependency Pairs:

TOP(mark(c)) -> TOP(ok(c))
TOP(mark(b)) -> TOP(ok(b))
TOP(mark(a)) -> TOP(ok(a))
TOP(mark(f(X1', X2', X3'))) -> TOP(f(proper(X1'), proper(X2'), proper(X3')))
TOP(ok(X)) -> TOP(active(X))

Rules:

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

Strategy:

innermost

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

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

five new Dependency Pairs are created:

TOP(ok(f(a, b, X''))) -> TOP(mark(f(X'', X'', X'')))
TOP(ok(c)) -> TOP(mark(a))
TOP(ok(c)) -> TOP(mark(b))
TOP(ok(f(X1', X2', X3'))) -> TOP(f(active(X1'), X2', X3'))
TOP(ok(f(X1', X2', X3'))) -> TOP(f(X1', X2', active(X3')))

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 10

             ↳Nar

...

               →DP Problem 11

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

TOP(ok(f(X1', X2', X3'))) -> TOP(f(X1', X2', active(X3')))
TOP(ok(f(X1', X2', X3'))) -> TOP(f(active(X1'), X2', X3'))
TOP(ok(f(a, b, X''))) -> TOP(mark(f(X'', X'', X'')))
TOP(mark(f(X1', X2', X3'))) -> TOP(f(proper(X1'), proper(X2'), proper(X3')))

Rules:

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

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:03 minutes

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

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

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