Term Rewriting System R:
[X, Y, Z, X1, X2]
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
AFROM(X) -> MARK(X)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(from(X)) -> AFROM(mark(X))
MARK(from(X)) -> MARK(X)
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)
MARK(from(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)


Strategy:

innermost




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

MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
12 new Dependency Pairs are created:

MARK(first(first(X1'', X2''), X2)) -> AFIRST(afirst(mark(X1''), mark(X2'')), mark(X2))
MARK(first(from(X'), X2)) -> AFIRST(afrom(mark(X')), mark(X2))
MARK(first(0, X2)) -> AFIRST(0, mark(X2))
MARK(first(nil, X2)) -> AFIRST(nil, mark(X2))
MARK(first(s(X'), X2)) -> AFIRST(s(mark(X')), mark(X2))
MARK(first(cons(X1'', X2''), X2)) -> AFIRST(cons(mark(X1''), X2''), mark(X2))
MARK(first(X1, first(X1'', X2''))) -> AFIRST(mark(X1), afirst(mark(X1''), mark(X2'')))
MARK(first(X1, from(X'))) -> AFIRST(mark(X1), afrom(mark(X')))
MARK(first(X1, 0)) -> AFIRST(mark(X1), 0)
MARK(first(X1, nil)) -> AFIRST(mark(X1), nil)
MARK(first(X1, s(X'))) -> AFIRST(mark(X1), s(mark(X')))
MARK(first(X1, cons(X1'', X2''))) -> AFIRST(mark(X1), cons(mark(X1''), X2''))

The transformation is resulting in one new DP problem: 



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




The following remains to be proven:
Dependency Pairs:

MARK(first(X1, cons(X1'', X2''))) -> AFIRST(mark(X1), cons(mark(X1''), X2''))
MARK(first(X1, from(X'))) -> AFIRST(mark(X1), afrom(mark(X')))
MARK(first(X1, first(X1'', X2''))) -> AFIRST(mark(X1), afirst(mark(X1''), mark(X2'')))
MARK(first(s(X'), X2)) -> AFIRST(s(mark(X')), mark(X2))
MARK(first(from(X'), X2)) -> AFIRST(afrom(mark(X')), mark(X2))
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(first(first(X1'', X2''), X2)) -> AFIRST(afirst(mark(X1''), mark(X2'')), mark(X2))
MARK(s(X)) -> MARK(X)
MARK(from(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(cons(X1, X2)) -> MARK(X1)


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:12 minutes