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

         ↳Size-Change Principle

       →DP Problem 2

         ↳Neg 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

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳SCP

       →DP Problem 2

         ↳Negative Polynomial Order

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 using the given order.

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

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

Used ordering:
Polynomial Order with Interpretation:

POL( ACTIVATE(x₁) ) = x₁

POL( n_{_diff}(x₁, x₂) ) = x₁ + x₂ + 1

POL( n_{_p}(x₁) ) = x₁

POL( DIFF(x₁, x₂) ) = x₁ + x₂ + 1

POL( activate(x₁) ) = x₁

POL( n_{_s}(x₁) ) = x₁

POL( IF(x₁, ..., x₃) ) = x₂ + x₃

POL( n_₀ ) = 0

POL( leq(x₁, x₂) ) = 0

POL( true ) = 0

POL( false ) = 0

POL( 0 ) = 0

POL( s(x₁) ) = x₁

POL( diff(x₁, x₂) ) = x₁ + x₂ + 1

POL( p(x₁) ) = x₁

POL( if(x₁, ..., x₃) ) = x₂ + x₃

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳SCP

       →DP Problem 2

         ↳Neg POLO

           →DP Problem 3

             ↳Negative Polynomial Order

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 using the given order.

ACTIVATE(n_{_p}(X)) -> ACTIVATE(X)

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

Used ordering:
Polynomial Order with Interpretation:

POL( ACTIVATE(x₁) ) = x₁

POL( n_{_p}(x₁) ) = x₁ + 1

POL( IF(x₁, ..., x₃) ) = x₂ + x₃

POL( DIFF(x₁, x₂) ) = 0

POL( n_₀ ) = 0

POL( n_{_s}(x₁) ) = x₁

POL( n_{_diff}(x₁, x₂) ) = 0

POL( leq(x₁, x₂) ) = 0

POL( true ) = 0

POL( false ) = 0

POL( activate(x₁) ) = x₁

POL( 0 ) = 0

POL( s(x₁) ) = x₁

POL( diff(x₁, x₂) ) = 0

POL( p(x₁) ) = x₁ + 1

POL( if(x₁, ..., x₃) ) = x₂ + x₃

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳SCP

       →DP Problem 2

         ↳Neg POLO

           →DP Problem 3

             ↳Neg POLO

...

               →DP Problem 4

                 ↳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

         ↳SCP

       →DP Problem 2

         ↳Neg POLO

           →DP Problem 3

             ↳Neg POLO

...

               →DP Problem 5

                 ↳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

         ↳SCP

       →DP Problem 2

         ↳Neg POLO

           →DP Problem 3

             ↳Neg POLO

...

               →DP Problem 6

                 ↳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

The Proof could not be continued due to a Timeout.
Termination of R could not be shown.
Duration:
1:00 minutes