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:
d1(x) -> e1(u1(x))
d1(u1(x)) -> c1(x)
c1(u1(x)) -> b1(x)
v1(e1(x)) -> x
b1(u1(x)) -> a1(e1(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
d1(x) -> e1(u1(x))
d1(u1(x)) -> c1(x)
c1(u1(x)) -> b1(x)
v1(e1(x)) -> x
b1(u1(x)) -> a1(e1(x))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
C1(u1(x)) -> B1(x)
D1(u1(x)) -> C1(x)
The TRS R consists of the following rules:
d1(x) -> e1(u1(x))
d1(u1(x)) -> c1(x)
c1(u1(x)) -> b1(x)
v1(e1(x)) -> x
b1(u1(x)) -> a1(e1(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:
C1(u1(x)) -> B1(x)
D1(u1(x)) -> C1(x)
The TRS R consists of the following rules:
d1(x) -> e1(u1(x))
d1(u1(x)) -> c1(x)
c1(u1(x)) -> b1(x)
v1(e1(x)) -> x
b1(u1(x)) -> a1(e1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.