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(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
activate(n__f(X)) → f(X)
activate(X) → X
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(activate(x1)) = 1 + 2·x1
POL(cons(x1, x2)) = x1 + 2·x2
POL(f(x1)) = 2·x1
POL(n__f(x1)) = 2·x1
POL(p(x1)) = 2·x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
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(0) → cons(0, n__f(s(0)))
f(X) → n__f(X)
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(cons(x1, x2)) = 1 + x1 + x2
POL(f(x1)) = 2 + x1
POL(n__f(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(0)) → f(p(s(0)))
p(s(0)) → 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:
p(s(0)) → 0
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(f(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 1 + 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(0)) → f(p(s(0)))
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(s(0)) → f(p(s(0)))
The signature Sigma is {f}
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(0)) → f(p(s(0)))
The set Q consists of the following terms:
f(s(0))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(s(0)) → F(p(s(0)))
The TRS R consists of the following rules:
f(s(0)) → f(p(s(0)))
The set Q consists of the following terms:
f(s(0))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(s(0)) → F(p(s(0)))
The TRS R consists of the following rules:
f(s(0)) → f(p(s(0)))
The set Q consists of the following terms:
f(s(0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.