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(a) → f(b)
g(b) → g(a)
f(x) → g(x)
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
f(x) → g(x)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
f(x) → g(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(x) → g(x)
Used ordering:
Polynomial interpretation [25]:
POL(a) = 0
POL(b) = 0
POL(f(x1)) = 2 + 2·x1
POL(g(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(a) → f(b)
g(b) → g(a)
The signature Sigma is {f, g}
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
The set Q consists of the following terms:
f(a)
g(b)
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(b) → G(a)
F(a) → F(b)
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
The set Q consists of the following terms:
f(a)
g(b)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
G(b) → G(a)
F(a) → F(b)
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
The set Q consists of the following terms:
f(a)
g(b)
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.