Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → X
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → CONS(X, n__from(s(X)))
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
2ND(cons(X, n__cons(Y, Z))) → ACTIVATE(Y)
The TRS R consists of the following rules:
2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → CONS(X, n__from(s(X)))
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
2ND(cons(X, n__cons(Y, Z))) → ACTIVATE(Y)
The TRS R consists of the following rules:
2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 4 less nodes.