Term Rewriting System R:
[X, XS, N, X1, X2]
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
aafter(0, XS) -> mark(XS)
aafter(s(N), cons(X, XS)) -> aafter(mark(N), mark(XS))
aafter(X1, X2) -> after(X1, X2)
mark(from(X)) -> afrom(mark(X))
mark(after(X1, X2)) -> aafter(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AFROM(X) -> MARK(X)
AAFTER(0, XS) -> MARK(XS)
AAFTER(s(N), cons(X, XS)) -> AAFTER(mark(N), mark(XS))
AAFTER(s(N), cons(X, XS)) -> MARK(N)
AAFTER(s(N), cons(X, XS)) -> MARK(XS)
MARK(from(X)) -> AFROM(mark(X))
MARK(from(X)) -> MARK(X)
MARK(after(X1, X2)) -> AAFTER(mark(X1), mark(X2))
MARK(after(X1, X2)) -> MARK(X1)
MARK(after(X1, X2)) -> MARK(X2)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains one SCC. 


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




The following remains to be proven:
Dependency Pairs:

AAFTER(s(N), cons(X, XS)) -> MARK(XS)
AAFTER(s(N), cons(X, XS)) -> MARK(N)
AAFTER(s(N), cons(X, XS)) -> AAFTER(mark(N), mark(XS))
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(after(X1, X2)) -> MARK(X2)
MARK(after(X1, X2)) -> MARK(X1)
AAFTER(0, XS) -> MARK(XS)
MARK(after(X1, X2)) -> AAFTER(mark(X1), mark(X2))
MARK(from(X)) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
AFROM(X) -> MARK(X)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
aafter(0, XS) -> mark(XS)
aafter(s(N), cons(X, XS)) -> aafter(mark(N), mark(XS))
aafter(X1, X2) -> after(X1, X2)
mark(from(X)) -> afrom(mark(X))
mark(after(X1, X2)) -> aafter(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0


Strategy:

innermost



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