Term Rewriting System R:
[X, Y]
f(X) -> if(X, c, nf(ntrue))
f(X) -> nf(X)
if(true, X, Y) -> X
if(false, X, Y) -> activate(Y)
true -> ntrue
activate(nf(X)) -> f(activate(X))
activate(ntrue) -> true
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(X) -> IF(X, c, nf(ntrue))
IF(false, X, Y) -> ACTIVATE(Y)
ACTIVATE(nf(X)) -> F(activate(X))
ACTIVATE(nf(X)) -> ACTIVATE(X)
ACTIVATE(ntrue) -> TRUE

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Instantiation Transformation`

Dependency Pairs:

ACTIVATE(nf(X)) -> ACTIVATE(X)
ACTIVATE(nf(X)) -> F(activate(X))
IF(false, X, Y) -> ACTIVATE(Y)
F(X) -> IF(X, c, nf(ntrue))

Rules:

f(X) -> if(X, c, nf(ntrue))
f(X) -> nf(X)
if(true, X, Y) -> X
if(false, X, Y) -> activate(Y)
true -> ntrue
activate(nf(X)) -> f(activate(X))
activate(ntrue) -> true
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, c, nf(ntrue)) -> ACTIVATE(nf(ntrue))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

IF(false, c, nf(ntrue)) -> ACTIVATE(nf(ntrue))
F(X) -> IF(X, c, nf(ntrue))
ACTIVATE(nf(X)) -> F(activate(X))
ACTIVATE(nf(X)) -> ACTIVATE(X)

Rules:

f(X) -> if(X, c, nf(ntrue))
f(X) -> nf(X)
if(true, X, Y) -> X
if(false, X, Y) -> activate(Y)
true -> ntrue
activate(nf(X)) -> f(activate(X))
activate(ntrue) -> true
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(nf(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(n__f(x1)) =  1 + x1 POL(n__true) =  0 POL(if(x1, x2, x3)) =  0 POL(c) =  1 POL(false) =  0 POL(true) =  0 POL(ACTIVATE(x1)) =  x1 POL(f(x1)) =  0 POL(IF(x1, x2, x3)) =  x2 POL(F(x1)) =  1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳Polo`
`             ...`
`               →DP Problem 3`
`                 ↳Instantiation Transformation`

Dependency Pairs:

IF(false, c, nf(ntrue)) -> ACTIVATE(nf(ntrue))
F(X) -> IF(X, c, nf(ntrue))
ACTIVATE(nf(X)) -> F(activate(X))

Rules:

f(X) -> if(X, c, nf(ntrue))
f(X) -> nf(X)
if(true, X, Y) -> X
if(false, X, Y) -> activate(Y)
true -> ntrue
activate(nf(X)) -> f(activate(X))
activate(ntrue) -> true
activate(X) -> X

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

ACTIVATE(nf(X)) -> F(activate(X))
one new Dependency Pair is created:

ACTIVATE(nf(ntrue)) -> F(activate(ntrue))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

F(X) -> IF(X, c, nf(ntrue))
ACTIVATE(nf(ntrue)) -> F(activate(ntrue))
IF(false, c, nf(ntrue)) -> ACTIVATE(nf(ntrue))

Rules:

f(X) -> if(X, c, nf(ntrue))
f(X) -> nf(X)
if(true, X, Y) -> X
if(false, X, Y) -> activate(Y)
true -> ntrue
activate(nf(X)) -> f(activate(X))
activate(ntrue) -> true
activate(X) -> X

The following dependency pair can be strictly oriented:

IF(false, c, nf(ntrue)) -> ACTIVATE(nf(ntrue))

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

activate(nf(X)) -> f(activate(X))
activate(ntrue) -> true
activate(X) -> X
if(true, X, Y) -> X
if(false, X, Y) -> activate(Y)
f(X) -> if(X, c, nf(ntrue))
f(X) -> nf(X)
true -> ntrue

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(activate(x1)) =  x1 POL(n__f(x1)) =  0 POL(n__true) =  0 POL(if(x1, x2, x3)) =  x2 + x3 POL(c) =  0 POL(false) =  1 POL(true) =  0 POL(ACTIVATE(x1)) =  0 POL(f(x1)) =  0 POL(F(x1)) =  x1 POL(IF(x1, x2, x3)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳Polo`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`

Dependency Pairs:

F(X) -> IF(X, c, nf(ntrue))
ACTIVATE(nf(ntrue)) -> F(activate(ntrue))

Rules:

f(X) -> if(X, c, nf(ntrue))
f(X) -> nf(X)
if(true, X, Y) -> X
if(false, X, Y) -> activate(Y)
true -> ntrue
activate(nf(X)) -> f(activate(X))
activate(ntrue) -> true
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes