YES
Termination w.r.t. Q proof of ../tpdb/TRS/nontermin/AG01/#4.20a.trs
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:
f(f(x)) → f(x)
f(s(x)) → f(x)
g(s(0)) → g(f(s(0)))
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(x)) → f(x)
f(s(x)) → f(x)
g(s(0)) → g(f(s(0)))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(x)) → f(x)
f(s(x)) → f(x)
g(s(0)) → g(f(s(0)))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
f(s(x)) → f(x)
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(f(x1)) = x1
POL(g(x1)) = x1
POL(s(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(x)) → f(x)
g(s(0)) → g(f(s(0)))
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:
G(s(0)) → F(s(0))
G(s(0)) → G(f(s(0)))
The TRS R consists of the following rules:
f(f(x)) → f(x)
g(s(0)) → g(f(s(0)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
G(s(0)) → F(s(0))
G(s(0)) → G(f(s(0)))
The TRS R consists of the following rules:
f(f(x)) → f(x)
g(s(0)) → g(f(s(0)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.