Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
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:
INCR(cons(X, XS)) → ACTIVATE(XS)
ODDS → INCR(pairs)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ODDS → PAIRS
ACTIVATE(n__nats) → NATS
TAIL(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__odds) → ODDS
The TRS R consists of the following rules:
nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
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:
INCR(cons(X, XS)) → ACTIVATE(XS)
ODDS → INCR(pairs)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ODDS → PAIRS
ACTIVATE(n__nats) → NATS
TAIL(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__odds) → ODDS
The TRS R consists of the following rules:
nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
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 1 SCC with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INCR(cons(X, XS)) → ACTIVATE(XS)
ODDS → INCR(pairs)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__odds) → ODDS
The TRS R consists of the following rules:
nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.