Termination Proof using AProVE

Term Rewriting System R:
[Y, X, X1, X2]
minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> n_{_minus}(X1, X2)
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}(n_{_div}(n_{_minus}(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
div(X1, X2) -> n_{_div}(X1, X2)
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(activate(X))
activate(n_{_div}(X1, X2)) -> div(activate(X1), X2)
activate(n_{_minus}(X1, X2)) -> minus(X1, X2)
activate(X) -> X

Innermost 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}(n_{_div}(n_{_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)
IF(true, X, Y) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
ACTIVATE(n_₀) -> 0'
ACTIVATE(n_{_s}(X)) -> S(activate(X))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_div}(X1, X2)) -> DIV(activate(X1), X2)
ACTIVATE(n_{_div}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_minus}(X1, X2)) -> MINUS(X1, X2)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

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))
minus(X1, X2) -> n_{_minus}(X1, X2)
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}(n_{_div}(n_{_minus}(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
div(X1, X2) -> n_{_div}(X1, X2)
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(activate(X))
activate(n_{_div}(X1, X2)) -> div(activate(X1), X2)
activate(n_{_minus}(X1, X2)) -> minus(X1, X2)
activate(X) -> X

Strategy:

innermost

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 for innermost can be oriented:

activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_div}(X1, X2)) -> div(activate(X1), X2)
activate(n_{_minus}(X1, X2)) -> minus(X1, X2)
activate(X) -> X
minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> n_{_minus}(X1, X2)
geq(X, n_₀) -> true
geq(n_₀, n_{_s}(Y)) -> false
geq(n_{_s}(X), n_{_s}(Y)) -> geq(activate(X), activate(Y))
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
div(0, n_{_s}(Y)) -> 0
div(s(X), n_{_s}(Y)) -> if(geq(X, activate(Y)), n_{_s}(n_{_div}(n_{_minus}(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
div(X1, X2) -> n_{_div}(X1, X2)
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(geq(x₁, x₂)) = 0

POL(false) = 0

POL(minus(x₁, x₂)) = 0

POL(n__s(x₁)) = 1 + x₁

POL(n__minus(x₁, x₂)) = 0

POL(true) = 0

POL(if(x₁, x₂, x₃)) = x₂ + x₃

POL(0) = 0

POL(n__div(x₁, x₂)) = x₁

POL(n__0) = 0

POL(s(x₁)) = 1 + x₁

POL(div(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> n_{_minus}(X1, X2)
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}(n_{_div}(n_{_minus}(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
div(X1, X2) -> n_{_div}(X1, X2)
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(activate(X))
activate(n_{_div}(X1, X2)) -> div(activate(X1), X2)
activate(n_{_minus}(X1, X2)) -> minus(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pairs:

MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(Y)
ACTIVATE(n_{_minus}(X1, X2)) -> MINUS(X1, X2)
ACTIVATE(n_{_div}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(X)
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))
minus(X1, X2) -> n_{_minus}(X1, X2)
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}(n_{_div}(n_{_minus}(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
div(X1, X2) -> n_{_div}(X1, X2)
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(activate(X))
activate(n_{_div}(X1, X2)) -> div(activate(X1), X2)
activate(n_{_minus}(X1, X2)) -> minus(X1, X2)
activate(X) -> X

Strategy:

innermost

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

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

10 new Dependency Pairs are created:

MINUS(n_{_s}(n_₀), n_{_s}(Y)) -> MINUS(0, activate(Y))
MINUS(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> MINUS(s(activate(X'')), activate(Y))
MINUS(n_{_s}(n_{_div}(X1', X2')), n_{_s}(Y)) -> MINUS(div(activate(X1'), X2'), activate(Y))
MINUS(n_{_s}(n_{_minus}(X1', X2')), n_{_s}(Y)) -> MINUS(minus(X1', X2'), activate(Y))
MINUS(n_{_s}(X''), n_{_s}(Y)) -> MINUS(X'', activate(Y))
MINUS(n_{_s}(X), n_{_s}(n_₀)) -> MINUS(activate(X), 0)
MINUS(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> MINUS(activate(X), s(activate(X'')))
MINUS(n_{_s}(X), n_{_s}(n_{_div}(X1', X2'))) -> MINUS(activate(X), div(activate(X1'), X2'))
MINUS(n_{_s}(X), n_{_s}(n_{_minus}(X1', X2'))) -> MINUS(activate(X), minus(X1', X2'))
MINUS(n_{_s}(X), n_{_s}(Y')) -> MINUS(activate(X), Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rewriting Transformation

Dependency Pairs:

MINUS(n_{_s}(X), n_{_s}(Y')) -> MINUS(activate(X), Y')
MINUS(n_{_s}(X), n_{_s}(n_{_minus}(X1', X2'))) -> MINUS(activate(X), minus(X1', X2'))
MINUS(n_{_s}(X), n_{_s}(n_{_div}(X1', X2'))) -> MINUS(activate(X), div(activate(X1'), X2'))
MINUS(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> MINUS(activate(X), s(activate(X'')))
MINUS(n_{_s}(X), n_{_s}(n_₀)) -> MINUS(activate(X), 0)
MINUS(n_{_s}(X''), n_{_s}(Y)) -> MINUS(X'', activate(Y))
MINUS(n_{_s}(n_{_minus}(X1', X2')), n_{_s}(Y)) -> MINUS(minus(X1', X2'), activate(Y))
MINUS(n_{_s}(n_{_div}(X1', X2')), n_{_s}(Y)) -> MINUS(div(activate(X1'), X2'), activate(Y))
MINUS(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> MINUS(s(activate(X'')), activate(Y))
MINUS(n_{_s}(n_₀), n_{_s}(Y)) -> MINUS(0, activate(Y))
MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(X)
ACTIVATE(n_{_minus}(X1, X2)) -> MINUS(X1, X2)
ACTIVATE(n_{_div}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(Y)

Rules:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> n_{_minus}(X1, X2)
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}(n_{_div}(n_{_minus}(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
div(X1, X2) -> n_{_div}(X1, X2)
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(activate(X))
activate(n_{_div}(X1, X2)) -> div(activate(X1), X2)
activate(n_{_minus}(X1, X2)) -> minus(X1, X2)
activate(X) -> X

Strategy:

innermost

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

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

one new Dependency Pair is created:

MINUS(n_{_s}(n_₀), n_{_s}(Y)) -> MINUS(n_₀, activate(Y))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 5

                 ↳Rewriting Transformation

Dependency Pairs:

MINUS(n_{_s}(X), n_{_s}(n_{_minus}(X1', X2'))) -> MINUS(activate(X), minus(X1', X2'))
MINUS(n_{_s}(X), n_{_s}(n_{_div}(X1', X2'))) -> MINUS(activate(X), div(activate(X1'), X2'))
MINUS(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> MINUS(activate(X), s(activate(X'')))
MINUS(n_{_s}(X), n_{_s}(n_₀)) -> MINUS(activate(X), 0)
MINUS(n_{_s}(X''), n_{_s}(Y)) -> MINUS(X'', activate(Y))
MINUS(n_{_s}(n_{_minus}(X1', X2')), n_{_s}(Y)) -> MINUS(minus(X1', X2'), activate(Y))
MINUS(n_{_s}(n_{_div}(X1', X2')), n_{_s}(Y)) -> MINUS(div(activate(X1'), X2'), activate(Y))
MINUS(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> MINUS(s(activate(X'')), activate(Y))
MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(Y)
ACTIVATE(n_{_minus}(X1, X2)) -> MINUS(X1, X2)
ACTIVATE(n_{_div}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(X)
MINUS(n_{_s}(X), n_{_s}(Y')) -> MINUS(activate(X), Y')

Rules:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> n_{_minus}(X1, X2)
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}(n_{_div}(n_{_minus}(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
div(X1, X2) -> n_{_div}(X1, X2)
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(activate(X))
activate(n_{_div}(X1, X2)) -> div(activate(X1), X2)
activate(n_{_minus}(X1, X2)) -> minus(X1, X2)
activate(X) -> X

Strategy:

innermost

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

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

one new Dependency Pair is created:

MINUS(n_{_s}(X), n_{_s}(n_₀)) -> MINUS(activate(X), n_₀)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 6

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MINUS(n_{_s}(X), n_{_s}(Y')) -> MINUS(activate(X), Y')
MINUS(n_{_s}(X), n_{_s}(n_{_div}(X1', X2'))) -> MINUS(activate(X), div(activate(X1'), X2'))
MINUS(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> MINUS(activate(X), s(activate(X'')))
MINUS(n_{_s}(X''), n_{_s}(Y)) -> MINUS(X'', activate(Y))
MINUS(n_{_s}(n_{_minus}(X1', X2')), n_{_s}(Y)) -> MINUS(minus(X1', X2'), activate(Y))
MINUS(n_{_s}(n_{_div}(X1', X2')), n_{_s}(Y)) -> MINUS(div(activate(X1'), X2'), activate(Y))
MINUS(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> MINUS(s(activate(X'')), activate(Y))
MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(Y)
ACTIVATE(n_{_minus}(X1, X2)) -> MINUS(X1, X2)
ACTIVATE(n_{_div}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
MINUS(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(X)
MINUS(n_{_s}(X), n_{_s}(n_{_minus}(X1', X2'))) -> MINUS(activate(X), minus(X1', X2'))

Rules:

minus(n_₀, Y) -> 0
minus(n_{_s}(X), n_{_s}(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> n_{_minus}(X1, X2)
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}(n_{_div}(n_{_minus}(X, activate(Y)), n_{_s}(activate(Y)))), n_₀)
div(X1, X2) -> n_{_div}(X1, X2)
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(activate(X))
activate(n_{_div}(X1, X2)) -> div(activate(X1), X2)
activate(n_{_minus}(X1, X2)) -> minus(X1, X2)
activate(X) -> X

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:20 minutes

POL(GEQ(x₁, x₂))	= 1 + x₁
POL(activate(x₁))	= x₁
POL(geq(x₁, x₂))	= 0
POL(false)	= 0
POL(minus(x₁, x₂))	= 0
POL(n__s(x₁))	= 1 + x₁
POL(n__minus(x₁, x₂))	= 0
POL(true)	= 0
POL(if(x₁, x₂, x₃))	= x₂ + x₃
POL(0)	= 0
POL(n__div(x₁, x₂))	= x₁
POL(n__0)	= 0
POL(s(x₁))	= 1 + x₁
POL(div(x₁, x₂))	= x₁