(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → 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:
FROM(X) → CONS(X, n__from(s(X)))
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH(n__cons(X, Y)) → ACTIVATE(Y)
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
The TRS R consists of the following rules:
from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → 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 6 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(X) → LENGTH(activate(X))
The TRS R consists of the following rules:
from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) NonLoopProof (EQUIVALENT transformation)
By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x1 / x0, x0 / s(x0)] on the rule
LENGTH(n__cons(x0, n__from(s(x0))))[ ]n[ ] → LENGTH(n__cons(x0, n__from(s(x0))))[ ]n[x1 / x0, x0 / s(x0)]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Simplify mu) - Instantiate mu
LENGTH(n__cons(x1, n__from(x0)))[ ]n[ ] → LENGTH(n__cons(x0, n__from(s(x0))))[ ]n[ ]
by Rewrite t
LENGTH(n__cons(x1, n__from(x0)))[ ]n[ ] → LENGTH(activate(from(x0)))[ ]n[ ]
by Narrowing at position: []
intermediate steps: Instantiation
LENGTH(n__cons(x1, n__from(y0)))[ ]n[ ] → LENGTH1(from(y0))[ ]n[ ]
by Narrowing at position: [0]
intermediate steps: Instantiation - Instantiation
LENGTH(n__cons(X, Y))[ ]n[ ] → LENGTH1(activate(Y))[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Instantiation
activate(n__from(X))[ ]n[ ] → from(X)[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Instantiation - Instantiation
LENGTH1(X)[ ]n[ ] → LENGTH(activate(X))[ ]n[ ]
by OriginalRule from TRS P
(6) NO