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:
a(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
b(x1) → a(x1)
c(c(x1)) → d(f(x1))
d(d(x1)) → f(f(f(x1)))
d(x1) → b(x1)
f(f(x1)) → g(a(x1))
g(g(x1)) → a(x1)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
b(x1) → a(x1)
c(c(x1)) → d(f(x1))
d(d(x1)) → f(f(f(x1)))
d(x1) → b(x1)
f(f(x1)) → g(a(x1))
g(g(x1)) → a(x1)
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:
D(x1) → B(x1)
F(f(x1)) → A(x1)
A(a(x1)) → C(x1)
G(g(x1)) → A(x1)
B(x1) → A(x1)
C(c(x1)) → F(x1)
B(b(x1)) → C(d(x1))
A(a(x1)) → B(c(x1))
B(b(x1)) → D(x1)
D(d(x1)) → F(f(f(x1)))
D(d(x1)) → F(x1)
F(f(x1)) → G(a(x1))
C(c(x1)) → D(f(x1))
D(d(x1)) → F(f(x1))
The TRS R consists of the following rules:
a(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
b(x1) → a(x1)
c(c(x1)) → d(f(x1))
d(d(x1)) → f(f(f(x1)))
d(x1) → b(x1)
f(f(x1)) → g(a(x1))
g(g(x1)) → a(x1)
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:
D(x1) → B(x1)
F(f(x1)) → A(x1)
A(a(x1)) → C(x1)
G(g(x1)) → A(x1)
B(x1) → A(x1)
C(c(x1)) → F(x1)
B(b(x1)) → C(d(x1))
A(a(x1)) → B(c(x1))
B(b(x1)) → D(x1)
D(d(x1)) → F(f(f(x1)))
D(d(x1)) → F(x1)
F(f(x1)) → G(a(x1))
C(c(x1)) → D(f(x1))
D(d(x1)) → F(f(x1))
The TRS R consists of the following rules:
a(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
b(x1) → a(x1)
c(c(x1)) → d(f(x1))
d(d(x1)) → f(f(f(x1)))
d(x1) → b(x1)
f(f(x1)) → g(a(x1))
g(g(x1)) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.