Termination Proof using AProVE

Term Rewriting System R:
[Y, X]
minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
geq(X, n_₀) -> true
geq(n_₀, n_{_s}(Y)) -> false
geq(n_{_s}(X), n_{_s}(Y)) -> geq(activate(X), activate(Y))
div(0, n_{_s}(Y)) -> 0
div(s(X), n_{_s}(Y)) -> if(geq(X, activate(Y)), n_{_s}(div(minus(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(n_₀, Y) -> 0'
MINUS(n_{_s}(X), n_{_s}(Y)) -> MINUS(activate(X), activate(Y))
MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(X)
MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(Y)
GEQ(n_{_s}(X), n_{_s}(Y)) -> GEQ(activate(X), activate(Y))
GEQ(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(X)
GEQ(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(Y)
DIV(s(X), n_{_s}(Y)) -> IF(geq(X, activate(Y)), n_{_s}(div(minus(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
DIV(s(X), n_{_s}(Y)) -> GEQ(X, activate(Y))
DIV(s(X), n_{_s}(Y)) -> ACTIVATE(Y)
DIV(s(X), n_{_s}(Y)) -> DIV(minus(X, activate(Y)), n_{_s}(activate(Y)))
DIV(s(X), n_{_s}(Y)) -> MINUS(X, activate(Y))
IF(true, X, Y) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
ACTIVATE(n_₀) -> 0'
ACTIVATE(n_{_s}(X)) -> S(X)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

MINUS(n_{_s}(X), n_{_s}(Y)) -> MINUS(activate(X), activate(Y))

Rules:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
geq(X, n_₀) -> true
geq(n_₀, n_{_s}(Y)) -> false
geq(n_{_s}(X), n_{_s}(Y)) -> geq(activate(X), activate(Y))
div(0, n_{_s}(Y)) -> 0
div(s(X), n_{_s}(Y)) -> if(geq(X, activate(Y)), n_{_s}(div(minus(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(X) -> X

The following dependency pair can be strictly oriented:

MINUS(n_{_s}(X), n_{_s}(Y)) -> MINUS(activate(X), activate(Y))

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

activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(X) -> X
0 -> n_₀
s(X) -> n_{_s}(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(activate(x₁)) = x₁

POL(0) = 0

POL(MINUS(x₁, x₂)) = 1 + x₁

POL(n__s(x₁)) = 1 + x₁

POL(n__0) = 0

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:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
geq(X, n_₀) -> true
geq(n_₀, n_{_s}(Y)) -> false
geq(n_{_s}(X), n_{_s}(Y)) -> geq(activate(X), activate(Y))
div(0, n_{_s}(Y)) -> 0
div(s(X), n_{_s}(Y)) -> if(geq(X, activate(Y)), n_{_s}(div(minus(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(X) -> 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

Dependency Pair:

GEQ(n_{_s}(X), n_{_s}(Y)) -> GEQ(activate(X), activate(Y))

Rules:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
geq(X, n_₀) -> true
geq(n_₀, n_{_s}(Y)) -> false
geq(n_{_s}(X), n_{_s}(Y)) -> geq(activate(X), activate(Y))
div(0, n_{_s}(Y)) -> 0
div(s(X), n_{_s}(Y)) -> if(geq(X, activate(Y)), n_{_s}(div(minus(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(X) -> X

The following dependency pair can be strictly oriented:

GEQ(n_{_s}(X), n_{_s}(Y)) -> GEQ(activate(X), activate(Y))

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

activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(X) -> X
0 -> n_₀
s(X) -> n_{_s}(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(GEQ(x₁, x₂)) = 1 + x₁

POL(activate(x₁)) = x₁

POL(0) = 0

POL(n__s(x₁)) = 1 + x₁

POL(n__0) = 0

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:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
geq(X, n_₀) -> true
geq(n_₀, n_{_s}(Y)) -> false
geq(n_{_s}(X), n_{_s}(Y)) -> geq(activate(X), activate(Y))
div(0, n_{_s}(Y)) -> 0
div(s(X), n_{_s}(Y)) -> if(geq(X, activate(Y)), n_{_s}(div(minus(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(X) -> 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

Dependency Pair:

DIV(s(X), n_{_s}(Y)) -> DIV(minus(X, activate(Y)), n_{_s}(activate(Y)))

Rules:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
geq(X, n_₀) -> true
geq(n_₀, n_{_s}(Y)) -> false
geq(n_{_s}(X), n_{_s}(Y)) -> geq(activate(X), activate(Y))
div(0, n_{_s}(Y)) -> 0
div(s(X), n_{_s}(Y)) -> if(geq(X, activate(Y)), n_{_s}(div(minus(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(X) -> X

The following dependency pair can be strictly oriented:

DIV(s(X), n_{_s}(Y)) -> DIV(minus(X, activate(Y)), n_{_s}(activate(Y)))

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

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
0 -> n_₀

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(activate(x₁)) = 0

POL(0) = 0

POL(DIV(x₁, x₂)) = x₁

POL(minus(x₁, x₂)) = 0

POL(n__s(x₁)) = 0

POL(n__0) = 0

POL(s(x₁)) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 6

             ↳Dependency Graph

Dependency Pair:

Rules:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
geq(X, n_₀) -> true
geq(n_₀, n_{_s}(Y)) -> false
geq(n_{_s}(X), n_{_s}(Y)) -> geq(activate(X), activate(Y))
div(0, n_{_s}(Y)) -> 0
div(s(X), n_{_s}(Y)) -> if(geq(X, activate(Y)), n_{_s}(div(minus(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes

POL(activate(x₁))	= x₁
POL(0)	= 0
POL(MINUS(x₁, x₂))	= 1 + x₁
POL(n__s(x₁))	= 1 + x₁
POL(n__0)	= 0
POL(s(x₁))	= 1 + x₁

POL(GEQ(x₁, x₂))	= 1 + x₁
POL(activate(x₁))	= x₁
POL(0)	= 0
POL(n__s(x₁))	= 1 + x₁
POL(n__0)	= 0
POL(s(x₁))	= 1 + x₁

POL(activate(x₁))	= 0
POL(0)	= 0
POL(DIV(x₁, x₂))	= x₁
POL(minus(x₁, x₂))	= 0
POL(n__s(x₁))	= 0
POL(n__0)	= 0
POL(s(x₁))	= 1