Termination Proof using AProVE

Term Rewriting System R:
[Y, X, X1, X2, X3]
a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_MINUS}(s(X), s(Y)) -> A_{_MINUS}(X, Y)
A_{_GEQ}(s(X), s(Y)) -> A_{_GEQ}(X, Y)
A_{_DIV}(s(X), s(Y)) -> A_{_IF}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
A_{_DIV}(s(X), s(Y)) -> A_{_GEQ}(X, Y)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(minus(X1, X2)) -> A_{_MINUS}(X1, X2)
MARK(geq(X1, X2)) -> A_{_GEQ}(X1, X2)
MARK(div(X1, X2)) -> A_{_DIV}(mark(X1), X2)
MARK(div(X1, X2)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

A_{_MINUS}(s(X), s(Y)) -> A_{_MINUS}(X, Y)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

The following dependency pair can be strictly oriented:

A_{_MINUS}(s(X), s(Y)) -> A_{_MINUS}(X, Y)

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(A__MINUS(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

Dependency Pair:

A_{_GEQ}(s(X), s(Y)) -> A_{_GEQ}(X, Y)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

The following dependency pair can be strictly oriented:

A_{_GEQ}(s(X), s(Y)) -> A_{_GEQ}(X, Y)

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(A__GEQ(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

Dependency Pair:

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(if(X1, X2, X3)) -> MARK(X1)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)
MARK(div(X1, X2)) -> MARK(X1)
MARK(div(X1, X2)) -> A_{_DIV}(mark(X1), X2)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_DIV}(s(X), s(Y)) -> A_{_IF}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

The following dependency pairs can be strictly oriented:

MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(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)) = x₁

POL(geq(x₁, x₂)) = 0

POL(false) = 0

POL(minus(x₁, x₂)) = 0

POL(true) = 0

POL(mark(x₁)) = 0

POL(a__div(x₁, x₂)) = 0

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

POL(0) = 0

POL(a__if(x₁, x₂, x₃)) = 0

POL(a__minus(x₁, x₂)) = 0

POL(s(x₁)) = x₁

POL(a__geq(x₁, x₂)) = 0

POL(div(x₁, x₂)) = x₁

POL(A__DIV(x₁, x₂)) = 0

POL(A__IF(x₁, x₂, x₃)) = x₂ + x₃

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 6

             ↳Polynomial Ordering

Dependency Pairs:

MARK(s(X)) -> MARK(X)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(div(X1, X2)) -> MARK(X1)
MARK(div(X1, X2)) -> A_{_DIV}(mark(X1), X2)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_DIV}(s(X), s(Y)) -> A_{_IF}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

The following dependency pair can be strictly oriented:

MARK(div(X1, X2)) -> MARK(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(MARK(x₁)) = x₁

POL(geq(x₁, x₂)) = 0

POL(false) = 0

POL(minus(x₁, x₂)) = 0

POL(true) = 0

POL(mark(x₁)) = 0

POL(a__div(x₁, x₂)) = 0

POL(if(x₁, x₂, x₃)) = 0

POL(0) = 0

POL(a__if(x₁, x₂, x₃)) = 0

POL(a__minus(x₁, x₂)) = 0

POL(s(x₁)) = x₁

POL(a__geq(x₁, x₂)) = 0

POL(div(x₁, x₂)) = 1 + x₁

POL(A__DIV(x₁, x₂)) = 1

POL(A__IF(x₁, x₂, x₃)) = x₂ + x₃

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

MARK(s(X)) -> MARK(X)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(div(X1, X2)) -> A_{_DIV}(mark(X1), X2)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_DIV}(s(X), s(Y)) -> A_{_IF}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

The following dependency pair can be strictly oriented:

MARK(s(X)) -> MARK(X)

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MARK(x₁)) = x₁

POL(geq(x₁, x₂)) = 0

POL(false) = 0

POL(minus(x₁, x₂)) = 0

POL(true) = 0

POL(mark(x₁)) = x₁

POL(a__div(x₁, x₂)) = x₁

POL(if(x₁, x₂, x₃)) = x₂ + x₃

POL(0) = 0

POL(a__if(x₁, x₂, x₃)) = x₂ + x₃

POL(a__minus(x₁, x₂)) = 0

POL(s(x₁)) = 1 + x₁

POL(a__geq(x₁, x₂)) = 0

POL(div(x₁, x₂)) = x₁

POL(A__DIV(x₁, x₂)) = x₁

POL(A__IF(x₁, x₂, x₃)) = x₂ + x₃

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 8

                 ↳Polynomial Ordering

Dependency Pairs:

A_{_IF}(false, X, Y) -> MARK(Y)
MARK(div(X1, X2)) -> A_{_DIV}(mark(X1), X2)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_DIV}(s(X), s(Y)) -> A_{_IF}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

The following dependency pair can be strictly oriented:

MARK(div(X1, X2)) -> A_{_DIV}(mark(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)) = x₁

POL(geq(x₁, x₂)) = 0

POL(false) = 0

POL(minus(x₁, x₂)) = 0

POL(true) = 0

POL(mark(x₁)) = 0

POL(a__div(x₁, x₂)) = 0

POL(if(x₁, x₂, x₃)) = 0

POL(0) = 0

POL(a__if(x₁, x₂, x₃)) = 0

POL(a__minus(x₁, x₂)) = 0

POL(s(x₁)) = 0

POL(a__geq(x₁, x₂)) = 0

POL(div(x₁, x₂)) = 1

POL(A__DIV(x₁, x₂)) = 0

POL(A__IF(x₁, x₂, x₃)) = x₂ + x₃

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 9

                 ↳Dependency Graph

Dependency Pairs:

A_{_IF}(false, X, Y) -> MARK(Y)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_DIV}(s(X), s(Y)) -> A_{_IF}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes

POL(A__MINUS(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁

POL(MARK(x₁))	= x₁
POL(geq(x₁, x₂))	= 0
POL(false)	= 0
POL(minus(x₁, x₂))	= 0
POL(true)	= 0
POL(mark(x₁))	= 0
POL(a__div(x₁, x₂))	= 0
POL(if(x₁, x₂, x₃))	= 1 + x₁ + x₂ + x₃
POL(0)	= 0
POL(a__if(x₁, x₂, x₃))	= 0
POL(a__minus(x₁, x₂))	= 0
POL(s(x₁))	= x₁
POL(a__geq(x₁, x₂))	= 0
POL(div(x₁, x₂))	= x₁
POL(A__DIV(x₁, x₂))	= 0
POL(A__IF(x₁, x₂, x₃))	= x₂ + x₃