Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A__SEL(0, cons(X, Y)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(cons(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
A__FROM(X) → MARK(X)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
A__SEL(0, cons(X, Y)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(cons(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
A__FROM(X) → MARK(X)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A__SEL(0, cons(X, Y)) → MARK(X)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(sel(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.