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

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}(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 two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

       →DP Problem 2

         ↳Nar

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

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))

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Rewriting Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

MINUS(n_{_s}(X), n_{_s}(Y')) -> MINUS(activate(X), Y')
MINUS(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> MINUS(activate(X), s(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_{_s}(X'')), n_{_s}(Y)) -> MINUS(s(X''), activate(Y))
MINUS(n_{_s}(n_₀), n_{_s}(Y)) -> MINUS(0, 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

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

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 4

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 5

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

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

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

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 6

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

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_{_s}(X''))) -> MINUS(activate(X), s(X''))

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 7

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

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(X'', activate(Y))

three new Dependency Pairs are created:

MINUS(n_{_s}(X''), n_{_s}(n_₀)) -> MINUS(X'', 0)
MINUS(n_{_s}(X''), n_{_s}(n_{_s}(X'))) -> MINUS(X'', s(X'))
MINUS(n_{_s}(X''), n_{_s}(Y')) -> MINUS(X'', Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 8

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

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

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(X'', 0)

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 9

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

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_{_s}(X'))) -> MINUS(X'', s(X'))

one new Dependency Pair is created:

MINUS(n_{_s}(X''), n_{_s}(n_{_s}(X'))) -> MINUS(X'', n_{_s}(X'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 10

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

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

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), Y')

three new Dependency Pairs are created:

MINUS(n_{_s}(n_₀), n_{_s}(Y')) -> MINUS(0, Y')
MINUS(n_{_s}(n_{_s}(X'')), n_{_s}(Y')) -> MINUS(s(X''), Y')
MINUS(n_{_s}(X''), n_{_s}(Y')) -> MINUS(X'', Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 11

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

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

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, Y')

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 12

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 13

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

Additionally, the following usable rules for innermost 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)) = x₂

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

         ↳Nar

           →DP Problem 3

             ↳Rw

...

               →DP Problem 14

                 ↳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))
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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Narrowing Transformation

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

Strategy:

innermost

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

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

six new Dependency Pairs are created:

GEQ(n_{_s}(n_₀), n_{_s}(Y)) -> GEQ(0, activate(Y))
GEQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> GEQ(s(X''), activate(Y))
GEQ(n_{_s}(X''), n_{_s}(Y)) -> GEQ(X'', activate(Y))
GEQ(n_{_s}(X), n_{_s}(n_₀)) -> GEQ(activate(X), 0)
GEQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> GEQ(activate(X), s(X''))
GEQ(n_{_s}(X), n_{_s}(Y')) -> GEQ(activate(X), Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rewriting Transformation

Dependency Pairs:

GEQ(n_{_s}(X), n_{_s}(Y')) -> GEQ(activate(X), Y')
GEQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> GEQ(activate(X), s(X''))
GEQ(n_{_s}(X), n_{_s}(n_₀)) -> GEQ(activate(X), 0)
GEQ(n_{_s}(X''), n_{_s}(Y)) -> GEQ(X'', activate(Y))
GEQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> GEQ(s(X''), activate(Y))
GEQ(n_{_s}(n_₀), n_{_s}(Y)) -> GEQ(0, 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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 16

                 ↳Rewriting Transformation

Dependency Pairs:

GEQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> GEQ(activate(X), s(X''))
GEQ(n_{_s}(X), n_{_s}(n_₀)) -> GEQ(activate(X), 0)
GEQ(n_{_s}(X''), n_{_s}(Y)) -> GEQ(X'', activate(Y))
GEQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> GEQ(s(X''), activate(Y))
GEQ(n_{_s}(X), n_{_s}(Y')) -> GEQ(activate(X), 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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 17

                 ↳Rewriting Transformation

Dependency Pairs:

GEQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> GEQ(n_{_s}(X''), activate(Y))
GEQ(n_{_s}(X), n_{_s}(Y')) -> GEQ(activate(X), Y')
GEQ(n_{_s}(X), n_{_s}(n_₀)) -> GEQ(activate(X), 0)
GEQ(n_{_s}(X''), n_{_s}(Y)) -> GEQ(X'', activate(Y))
GEQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> GEQ(activate(X), s(X''))

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 18

                 ↳Rewriting Transformation

Dependency Pairs:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 19

                 ↳Narrowing Transformation

Dependency Pairs:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

GEQ(n_{_s}(X''), n_{_s}(n_₀)) -> GEQ(X'', 0)
GEQ(n_{_s}(X''), n_{_s}(n_{_s}(X'))) -> GEQ(X'', s(X'))
GEQ(n_{_s}(X''), n_{_s}(Y')) -> GEQ(X'', Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 20

                 ↳Rewriting Transformation

Dependency Pairs:

GEQ(n_{_s}(X''), n_{_s}(Y')) -> GEQ(X'', Y')
GEQ(n_{_s}(X''), n_{_s}(n_{_s}(X'))) -> GEQ(X'', s(X'))
GEQ(n_{_s}(X''), n_{_s}(n_₀)) -> GEQ(X'', 0)
GEQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> GEQ(n_{_s}(X''), activate(Y))
GEQ(n_{_s}(X), n_{_s}(Y')) -> GEQ(activate(X), Y')
GEQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> GEQ(activate(X), n_{_s}(X''))

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

Strategy:

innermost

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

GEQ(n_{_s}(X''), n_{_s}(n_₀)) -> GEQ(X'', 0)

one new Dependency Pair is created:

GEQ(n_{_s}(X''), n_{_s}(n_₀)) -> GEQ(X'', n_₀)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 21

                 ↳Rewriting Transformation

Dependency Pairs:

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

Strategy:

innermost

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

GEQ(n_{_s}(X''), n_{_s}(n_{_s}(X'))) -> GEQ(X'', s(X'))

one new Dependency Pair is created:

GEQ(n_{_s}(X''), n_{_s}(n_{_s}(X'))) -> GEQ(X'', n_{_s}(X'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 22

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

GEQ(n_{_s}(n_₀), n_{_s}(Y')) -> GEQ(0, Y')
GEQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y')) -> GEQ(s(X''), Y')
GEQ(n_{_s}(X''), n_{_s}(Y')) -> GEQ(X'', Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 23

                 ↳Rewriting Transformation

Dependency Pairs:

GEQ(n_{_s}(X''), n_{_s}(Y')) -> GEQ(X'', Y')
GEQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y')) -> GEQ(s(X''), Y')
GEQ(n_{_s}(n_₀), n_{_s}(Y')) -> GEQ(0, Y')
GEQ(n_{_s}(X''), n_{_s}(Y')) -> GEQ(X'', Y')
GEQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> GEQ(activate(X), n_{_s}(X''))
GEQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> GEQ(n_{_s}(X''), activate(Y))
GEQ(n_{_s}(X''), n_{_s}(n_{_s}(X'))) -> GEQ(X'', n_{_s}(X'))

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

Strategy:

innermost

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

GEQ(n_{_s}(n_₀), n_{_s}(Y')) -> GEQ(0, Y')

one new Dependency Pair is created:

GEQ(n_{_s}(n_₀), n_{_s}(Y')) -> GEQ(n_₀, Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 24

                 ↳Rewriting Transformation

Dependency Pairs:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 25

                 ↳Polynomial Ordering

Dependency Pairs:

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

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

Additionally, the following usable rules for innermost 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)) = x₂

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

         ↳Nar

       →DP Problem 2

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 26

                 ↳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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:05 minutes

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

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