Term Rewriting System R:
[X, Y, X1, X2]
p(0) -> 0
p(s(X)) -> X
p(X) -> np(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), n0, ns(ndiff(np(X), Y)))
diff(X1, X2) -> ndiff(X1, X2)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiff(X1, X2)) -> diff(activate(X1), activate(X2))
activate(np(X)) -> p(activate(X))
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳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), n0, ns(ndiff(np(X), Y)))
DIFF(X, Y) -> LEQ(X, Y)
ACTIVATE(n0) -> 0'
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(ndiff(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(ndiff(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ndiff(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(np(X)) -> P(activate(X))
ACTIVATE(np(X)) -> ACTIVATE(X)

Furthermore, R contains two SCCs.

`   R`
`     ↳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) -> np(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), n0, ns(ndiff(np(X), Y)))
diff(X1, X2) -> ndiff(X1, X2)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiff(X1, X2)) -> diff(activate(X1), activate(X2))
activate(np(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.

`   R`
`     ↳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) -> np(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), n0, ns(ndiff(np(X), Y)))
diff(X1, X2) -> ndiff(X1, X2)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiff(X1, X2)) -> diff(activate(X1), activate(X2))
activate(np(X)) -> p(activate(X))
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`

Dependency Pairs:

ACTIVATE(np(X)) -> ACTIVATE(X)
ACTIVATE(ndiff(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(ndiff(X1, X2)) -> ACTIVATE(X1)
IF(false, X, Y) -> ACTIVATE(Y)
DIFF(X, Y) -> IF(leq(X, Y), n0, ns(ndiff(np(X), Y)))
ACTIVATE(ndiff(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(ns(X)) -> ACTIVATE(X)
IF(true, X, Y) -> ACTIVATE(X)

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> np(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), n0, ns(ndiff(np(X), Y)))
diff(X1, X2) -> ndiff(X1, X2)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiff(X1, X2)) -> diff(activate(X1), activate(X2))
activate(np(X)) -> p(activate(X))
activate(X) -> X

The following dependency pairs can be strictly oriented:

ACTIVATE(ndiff(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(ndiff(X1, X2)) -> ACTIVATE(X1)

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

activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiff(X1, X2)) -> diff(activate(X1), activate(X2))
activate(np(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), n0, ns(ndiff(np(X), Y)))
diff(X1, X2) -> ndiff(X1, X2)
0 -> n0
s(X) -> ns(X)
p(0) -> 0
p(s(X)) -> X
p(X) -> np(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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Polynomial Ordering`

Dependency Pairs:

ACTIVATE(np(X)) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
DIFF(X, Y) -> IF(leq(X, Y), n0, ns(ndiff(np(X), Y)))
ACTIVATE(ndiff(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(ns(X)) -> ACTIVATE(X)
IF(true, X, Y) -> ACTIVATE(X)

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> np(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), n0, ns(ndiff(np(X), Y)))
diff(X1, X2) -> ndiff(X1, X2)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiff(X1, X2)) -> diff(activate(X1), activate(X2))
activate(np(X)) -> p(activate(X))
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(np(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.

`   R`
`     ↳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), n0, ns(ndiff(np(X), Y)))
ACTIVATE(ndiff(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(ns(X)) -> ACTIVATE(X)
IF(true, X, Y) -> ACTIVATE(X)

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> np(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), n0, ns(ndiff(np(X), Y)))
diff(X1, X2) -> ndiff(X1, X2)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiff(X1, X2)) -> diff(activate(X1), activate(X2))
activate(np(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, n0, ns(ndiff(np(X''), Y''))) -> ACTIVATE(n0)

The transformation is resulting in one new DP problem:

`   R`
`     ↳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), n0, ns(ndiff(np(X), Y)))
ACTIVATE(ndiff(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(ns(X)) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> np(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), n0, ns(ndiff(np(X), Y)))
diff(X1, X2) -> ndiff(X1, X2)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiff(X1, X2)) -> diff(activate(X1), activate(X2))
activate(np(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, n0, ns(ndiff(np(X''), Y''))) -> ACTIVATE(ns(ndiff(np(X''), Y'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳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(ndiff(X1, X2)) -> DIFF(activate(X1), activate(X2))
ACTIVATE(ns(X)) -> ACTIVATE(X)
IF(false, n0, ns(ndiff(np(X''), Y''))) -> ACTIVATE(ns(ndiff(np(X''), Y'')))
DIFF(X, Y) -> IF(leq(X, Y), n0, ns(ndiff(np(X), Y)))

Rules:

p(0) -> 0
p(s(X)) -> X
p(X) -> np(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), n0, ns(ndiff(np(X), Y)))
diff(X1, X2) -> ndiff(X1, X2)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiff(X1, X2)) -> diff(activate(X1), activate(X2))
activate(np(X)) -> p(activate(X))
activate(X) -> X

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