(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(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:
2ND(cons(X, X1)) → 2ND(cons1(X, X1))
FROM(X) → FROM(s(X))
The TRS R consists of the following rules:
2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(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:
FROM(X) → FROM(s(X))
The TRS R consists of the following rules:
2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(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' = [x0 / s(x0)] on the rule
FROM(s(x0))[ ]n[ ] → FROM(s(x0))[ ]n[x0 / s(x0)]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Simplify mu) - Instantiate mu - Instantiation
FROM(X)[ ]n[ ] → FROM(s(X))[ ]n[ ]
by OriginalRule from TRS P
(6) NO