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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
FIRST(x1, x2) -> FIRST(x1, x2)
s(x1) -> s(x1)
cons(x1, x2) -> cons(x1, x2)


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


Dependency Pair:


Rules:


first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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




The following remains to be proven:
Dependency Pair:

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


Rules:


first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))


Strategy:

innermost



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