Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2, X3]
active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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)) -> IF(X, c, f(true))
ACTIVE(f(X)) -> F(true)
ACTIVE(f(X)) -> F(active(X))
ACTIVE(f(X)) -> ACTIVE(X)
ACTIVE(if(X1, X2, X3)) -> IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) -> IF(X1, active(X2), X3)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X2)
F(mark(X)) -> F(X)
F(ok(X)) -> F(X)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)
IF(X1, mark(X2), X3) -> IF(X1, X2, X3)
IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
PROPER(f(X)) -> F(proper(X))
PROPER(f(X)) -> PROPER(X)
PROPER(if(X1, X2, X3)) -> IF(proper(X1), proper(X2), proper(X3))
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(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 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:

IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
IF(X1, mark(X2), X3) -> IF(X1, X2, X3)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

IF(ok(X1), ok(X2), ok(X3)) -> IF(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(mark(x₁)) = 0

POL(ok(x₁)) = 1 + x₁

POL(IF(x₁, x₂, x₃)) = x₃

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 Pairs:

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

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

IF(mark(X1), X2, X3) -> IF(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(mark(x₁)) = 1 + x₁

POL(IF(x₁, x₂, x₃)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 7

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

IF(X1, mark(X2), X3) -> IF(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(mark(x₁)) = 1 + x₁

POL(IF(x₁, x₂, x₃)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

F(ok(X)) -> F(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(mark(x₁)) = x₁

POL(ok(x₁)) = 1 + x₁

POL(F(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 9

             ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

F(mark(X)) -> F(X)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

F(mark(X)) -> F(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(mark(x₁)) = 1 + x₁

POL(F(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 9

             ↳Polo

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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(if(X1, X2, X3)) -> ACTIVE(X2)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

ACTIVE(if(X1, X2, X3)) -> ACTIVE(X2)
ACTIVE(if(X1, X2, X3)) -> 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(if(x₁, x₂, x₃)) = 1 + x₁ + x₂

POL(f(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 11

             ↳Polynomial Ordering

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

ACTIVE(f(X)) -> ACTIVE(X)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(f(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(f(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 11

             ↳Polo

...

               →DP Problem 12

                 ↳Dependency Graph

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(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(if(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃

POL(PROPER(x₁)) = x₁

POL(f(x₁)) = 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 13

             ↳Polynomial Ordering

       →DP Problem 5

         ↳Remaining

Dependency Pair:

PROPER(f(X)) -> PROPER(X)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

PROPER(f(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(f(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 13

             ↳Polo

...

               →DP Problem 14

                 ↳Dependency Graph

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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(IF(x₁, x₂, x₃))	= x₃

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

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

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