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:
f1(X) -> if3(X, c, n__f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(X) -> if3(X, c, n__f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__f1(X)) -> F1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
F1(X) -> IF3(X, c, n__f1(true))
The TRS R consists of the following rules:
f1(X) -> if3(X, c, n__f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__f1(X)) -> F1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
F1(X) -> IF3(X, c, n__f1(true))
The TRS R consists of the following rules:
f1(X) -> if3(X, c, n__f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.