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:
zeros → cons(0, n__zeros)
tail(cons(X, XS)) → activate(XS)
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
tail(cons(X, XS)) → activate(XS)
zeros → n__zeros
activate(n__zeros) → zeros
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:
TAIL(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zeros) → ZEROS
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
tail(cons(X, XS)) → activate(XS)
zeros → n__zeros
activate(n__zeros) → zeros
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:
TAIL(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zeros) → ZEROS
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
tail(cons(X, XS)) → activate(XS)
zeros → n__zeros
activate(n__zeros) → zeros
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 2 less nodes.