Term Rewriting System R:
[X, Y, Z]
and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

ADD(s(X), Y) -> ADD(X, Y)
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)
FROM(X) -> FROM(s(X))

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Remaining


Dependency Pair:

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


Rules:


and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{nil, first}
add > s

resulting in one new DP problem.
Used Argument Filtering System:
ADD(x1, x2) -> ADD(x1, x2)
s(x1) -> s(x1)
and(x1, x2) -> and(x1, x2)
if(x1, x2, x3) -> if(x1, x2, x3)
add(x1, x2) -> add(x1, x2)
first(x1, x2) -> first(x1, x2)
cons(x1, x2) -> x1
from(x1) -> from(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Remaining


Dependency Pair:


Rules:


and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳Remaining


Dependency Pair:

FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)


Rules:


and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))





The following dependency pair can be strictly oriented:

FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)


The following rules can be oriented:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{nil, first}
add > s

resulting in one new DP problem.
Used Argument Filtering System:
FIRST(x1, x2) -> FIRST(x1, x2)
s(x1) -> s(x1)
cons(x1, x2) -> x2
and(x1, x2) -> and(x1, x2)
if(x1, x2, x3) -> if(x1, x2, x3)
add(x1, x2) -> add(x1, x2)
first(x1, x2) -> first(x1, x2)
from(x1) -> from


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳Remaining


Dependency Pair:


Rules:


and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

FROM(X) -> FROM(s(X))


Rules:


and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))




Termination of R could not be shown.
Duration:
0:00 minutes