Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2]
p(0) -> 0
p(s(X)) -> X
p(X) -> n_{_p}(X)
leq(0, Y) -> true
leq(s(X), 0) -> false
leq(s(X), s(Y)) -> leq(X, Y)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
diff(X, Y) -> if(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
diff(X1, X2) -> n_{_diff}(X1, X2)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_diff}(X1, X2)) -> diff(activate(X1), activate(X2))
activate(n_{_p}(X)) -> p(activate(X))
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LEQ(s(X), s(Y)) -> LEQ(X, Y)
IF(true, X, Y) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
DIFF(X, Y) -> IF(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
DIFF(X, Y) -> LEQ(X, Y)
ACTIVATE(n_₀) -> 0'
ACTIVATE(n_{_s}(X)) -> S(activate(X))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_diff}(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(n_{_diff}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_diff}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_p}(X)) -> P(activate(X))
ACTIVATE(n_{_p}(X)) -> ACTIVATE(X)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

Dependency Pair:

LEQ(s(X), s(Y)) -> LEQ(X, Y)

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> n_{_p}(X)
leq(0, Y) -> true
leq(s(X), 0) -> false
leq(s(X), s(Y)) -> leq(X, Y)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
diff(X, Y) -> if(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
diff(X1, X2) -> n_{_diff}(X1, X2)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_diff}(X1, X2)) -> diff(activate(X1), activate(X2))
activate(n_{_p}(X)) -> p(activate(X))
activate(X) -> X

The following dependency pair can be strictly oriented:

LEQ(s(X), s(Y)) -> LEQ(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(LEQ(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

Dependency Pair:

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> n_{_p}(X)
leq(0, Y) -> true
leq(s(X), 0) -> false
leq(s(X), s(Y)) -> leq(X, Y)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
diff(X, Y) -> if(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
diff(X1, X2) -> n_{_diff}(X1, X2)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_diff}(X1, X2)) -> diff(activate(X1), activate(X2))
activate(n_{_p}(X)) -> p(activate(X))
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

Dependency Pairs:

ACTIVATE(n_{_p}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_diff}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_diff}(X1, X2)) -> ACTIVATE(X1)
IF(false, X, Y) -> ACTIVATE(Y)
DIFF(X, Y) -> IF(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
ACTIVATE(n_{_diff}(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
IF(true, X, Y) -> ACTIVATE(X)

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> n_{_p}(X)
leq(0, Y) -> true
leq(s(X), 0) -> false
leq(s(X), s(Y)) -> leq(X, Y)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
diff(X, Y) -> if(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
diff(X1, X2) -> n_{_diff}(X1, X2)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_diff}(X1, X2)) -> diff(activate(X1), activate(X2))
activate(n_{_p}(X)) -> p(activate(X))
activate(X) -> X

The following dependency pairs can be strictly oriented:

ACTIVATE(n_{_diff}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_diff}(X1, X2)) -> ACTIVATE(X1)

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

activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_diff}(X1, X2)) -> diff(activate(X1), activate(X2))
activate(n_{_p}(X)) -> p(activate(X))
activate(X) -> X
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
diff(X, Y) -> if(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
diff(X1, X2) -> n_{_diff}(X1, X2)
0 -> n_₀
s(X) -> n_{_s}(X)
p(0) -> 0
p(s(X)) -> X
p(X) -> n_{_p}(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(activate(x₁)) = x₁

POL(false) = 0

POL(true) = 0

POL(n__s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

POL(IF(x₁, x₂, x₃)) = x₂ + x₃

POL(DIFF(x₁, x₂)) = 1 + x₁ + x₂

POL(if(x₁, x₂, x₃)) = x₂ + x₃

POL(0) = 0

POL(n__diff(x₁, x₂)) = 1 + x₁ + x₂

POL(leq(x₁, x₂)) = 0

POL(s(x₁)) = x₁

POL(n__0) = 0

POL(diff(x₁, x₂)) = 1 + x₁ + x₂

POL(p(x₁)) = x₁

POL(n__p(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 4

             ↳Polynomial Ordering

Dependency Pairs:

ACTIVATE(n_{_p}(X)) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
DIFF(X, Y) -> IF(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
ACTIVATE(n_{_diff}(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
IF(true, X, Y) -> ACTIVATE(X)

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> n_{_p}(X)
leq(0, Y) -> true
leq(s(X), 0) -> false
leq(s(X), s(Y)) -> leq(X, Y)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
diff(X, Y) -> if(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
diff(X1, X2) -> n_{_diff}(X1, X2)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_diff}(X1, X2)) -> diff(activate(X1), activate(X2))
activate(n_{_p}(X)) -> p(activate(X))
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_p}(X)) -> ACTIVATE(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(activate(x₁)) = 0

POL(false) = 0

POL(true) = 0

POL(n__s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

POL(IF(x₁, x₂, x₃)) = x₂ + x₃

POL(DIFF(x₁, x₂)) = 0

POL(if(x₁, x₂, x₃)) = 0

POL(0) = 0

POL(n__diff(x₁, x₂)) = 0

POL(leq(x₁, x₂)) = 0

POL(s(x₁)) = 0

POL(n__0) = 0

POL(diff(x₁, x₂)) = 0

POL(p(x₁)) = 0

POL(n__p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 4

             ↳Polo

...

               →DP Problem 5

                 ↳Instantiation Transformation

Dependency Pairs:

IF(false, X, Y) -> ACTIVATE(Y)
DIFF(X, Y) -> IF(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
ACTIVATE(n_{_diff}(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
IF(true, X, Y) -> ACTIVATE(X)

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> n_{_p}(X)
leq(0, Y) -> true
leq(s(X), 0) -> false
leq(s(X), s(Y)) -> leq(X, Y)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
diff(X, Y) -> if(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
diff(X1, X2) -> n_{_diff}(X1, X2)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_diff}(X1, X2)) -> diff(activate(X1), activate(X2))
activate(n_{_p}(X)) -> p(activate(X))
activate(X) -> X

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

IF(true, X, Y) -> ACTIVATE(X)

one new Dependency Pair is created:

IF(true, n_₀, n_{_s}(n_{_diff}(n_{_p}(X''), Y''))) -> ACTIVATE(n_₀)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 4

             ↳Polo

...

               →DP Problem 6

                 ↳Instantiation Transformation

Dependency Pairs:

DIFF(X, Y) -> IF(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
ACTIVATE(n_{_diff}(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> n_{_p}(X)
leq(0, Y) -> true
leq(s(X), 0) -> false
leq(s(X), s(Y)) -> leq(X, Y)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
diff(X, Y) -> if(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
diff(X1, X2) -> n_{_diff}(X1, X2)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_diff}(X1, X2)) -> diff(activate(X1), activate(X2))
activate(n_{_p}(X)) -> p(activate(X))
activate(X) -> X

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

IF(false, X, Y) -> ACTIVATE(Y)

one new Dependency Pair is created:

IF(false, n_₀, n_{_s}(n_{_diff}(n_{_p}(X''), Y''))) -> ACTIVATE(n_{_s}(n_{_diff}(n_{_p}(X''), Y'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 4

             ↳Polo

...

               →DP Problem 7

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

ACTIVATE(n_{_diff}(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
IF(false, n_₀, n_{_s}(n_{_diff}(n_{_p}(X''), Y''))) -> ACTIVATE(n_{_s}(n_{_diff}(n_{_p}(X''), Y'')))
DIFF(X, Y) -> IF(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> n_{_p}(X)
leq(0, Y) -> true
leq(s(X), 0) -> false
leq(s(X), s(Y)) -> leq(X, Y)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
diff(X, Y) -> if(leq(X, Y), n_₀, n_{_s}(n_{_diff}(n_{_p}(X), Y)))
diff(X1, X2) -> n_{_diff}(X1, X2)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_diff}(X1, X2)) -> diff(activate(X1), activate(X2))
activate(n_{_p}(X)) -> p(activate(X))
activate(X) -> X

Termination of R could not be shown.
Duration:
0:01 minutes

POL(activate(x₁))	= x₁
POL(false)	= 0
POL(true)	= 0
POL(n__s(x₁))	= x₁
POL(ACTIVATE(x₁))	= x₁
POL(IF(x₁, x₂, x₃))	= x₂ + x₃
POL(DIFF(x₁, x₂))	= 1 + x₁ + x₂
POL(if(x₁, x₂, x₃))	= x₂ + x₃
POL(0)	= 0
POL(n__diff(x₁, x₂))	= 1 + x₁ + x₂
POL(leq(x₁, x₂))	= 0
POL(s(x₁))	= x₁
POL(n__0)	= 0
POL(diff(x₁, x₂))	= 1 + x₁ + x₂
POL(p(x₁))	= x₁
POL(n__p(x₁))	= x₁