Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
Q is empty.
↳ QTRS
↳ DirectTerminationProof
Q restricted rewrite system:
The TRS R consists of the following rules:
from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
Q is empty.
We use [23] with the following order to prove termination.
Lexicographic path order with status [19].
Quasi-Precedence:
2nd1 > head1
2nd1 > [activate1, take2] > from1 > [cons2, ntake2]
2nd1 > [activate1, take2] > from1 > nfrom1
2nd1 > [activate1, take2] > from1 > s1
[0, nil]
sel2 > [activate1, take2] > from1 > [cons2, ntake2]
sel2 > [activate1, take2] > from1 > nfrom1
sel2 > [activate1, take2] > from1 > s1
Status: from1: [1]
sel2: [1,2]
nfrom1: [1]
head1: [1]
0: multiset
nil: multiset
cons2: [1,2]
activate1: [1]
s1: [1]
2nd1: [1]
take2: [2,1]
ntake2: [2,1]