(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__zeros → cons(0, zeros)
a__tail(cons(X, XS)) → mark(XS)
mark(zeros) → a__zeros
mark(tail(X)) → a__tail(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
a__zeros → zeros
a__tail(X) → tail(X)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A__TAIL(cons(X, XS)) → MARK(XS)
MARK(zeros) → A__ZEROS
MARK(tail(X)) → A__TAIL(mark(X))
MARK(tail(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
a__zeros → cons(0, zeros)
a__tail(cons(X, XS)) → mark(XS)
mark(zeros) → a__zeros
mark(tail(X)) → a__tail(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
a__zeros → zeros
a__tail(X) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, XS)) → MARK(XS)
MARK(tail(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
a__zeros → cons(0, zeros)
a__tail(cons(X, XS)) → mark(XS)
mark(zeros) → a__zeros
mark(tail(X)) → a__tail(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
a__zeros → zeros
a__tail(X) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.