(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following 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))
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
length(nil) → 0
length(cons(X, L)) → s(length(L))
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
The TRS R 2 is
eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
The signature Sigma is {
eq,
true,
false}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following 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))
The set Q consists of the following terms:
eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
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)
The TRS R consists of the following 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))
The set Q consists of the following terms:
eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(X, L)) → LENGTH(L)
The TRS R consists of the following 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))
The set Q consists of the following terms:
eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TAKE(s(X), cons(Y, L)) → TAKE(X, L)
The TRS R consists of the following 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))
The set Q consists of the following terms:
eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INF(X) → INF(s(X))
The TRS R consists of the following 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))
The set Q consists of the following terms:
eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQ(s(X), s(Y)) → EQ(X, Y)
The TRS R consists of the following 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))
The set Q consists of the following terms:
eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.