(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__length(nil) → 0
a__length(cons(N, L)) → a__U11(tt, L)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(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__U11(tt, L) → A__U12(tt, L)
A__U12(tt, L) → A__LENGTH(mark(L))
A__U12(tt, L) → MARK(L)
A__LENGTH(cons(N, L)) → A__U11(tt, L)
MARK(zeros) → A__ZEROS
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U11(X1, X2)) → MARK(X1)
MARK(U12(X1, X2)) → A__U12(mark(X1), X2)
MARK(U12(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
MARK(length(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__length(nil) → 0
a__length(cons(N, L)) → a__U11(tt, L)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(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:
A__U12(tt, L) → A__LENGTH(mark(L))
A__LENGTH(cons(N, L)) → A__U11(tt, L)
A__U11(tt, L) → A__U12(tt, L)
A__U12(tt, L) → MARK(L)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U11(X1, X2)) → MARK(X1)
MARK(U12(X1, X2)) → A__U12(mark(X1), X2)
MARK(U12(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
MARK(length(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__length(nil) → 0
a__length(cons(N, L)) → a__U11(tt, L)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.