Term Rewriting System R:
[X, Y, L]
eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

EQ(s(X), s(Y)) -> EQ(X, Y)
INF(X) -> INF(s(X))
TAKE(s(X), cons(Y, L)) -> TAKE(X, L)
LENGTH(cons(X, L)) -> LENGTH(L)

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pair:

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


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost 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:
EQ(x1, x2) -> EQ(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 5
Dependency Graph
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pair:


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:

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