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