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