Termination w.r.t. Q of the following Term Rewriting System could be disproven:

Q restricted rewrite system:
The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
a → b

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

Q restricted rewrite system:
The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
a → b

Q is empty.

The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
a → b

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

a → b
Used ordering:
Polynomial interpretation [25]:

POL(a) = 1   
POL(b) = 0   
POL(f(x1)) = 1 + x1   
POL(g(x1)) = x1   
POL(h(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost

Q restricted rewrite system:
The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)

The signature Sigma is {f, g, h}

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)

The set Q consists of the following terms:

h(x0)
g(a)
f(x0)


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(a) → F(b)
H(X) → G(X)
F(X) → H(a)

The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)

The set Q consists of the following terms:

h(x0)
g(a)
f(x0)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
              ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

G(a) → F(b)
H(X) → G(X)
F(X) → H(a)

The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)

The set Q consists of the following terms:

h(x0)
g(a)
f(x0)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
              ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

G(a) → F(b)
H(X) → G(X)
F(X) → H(a)

R is empty.
The set Q consists of the following terms:

h(x0)
g(a)
f(x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

h(x0)
g(a)
f(x0)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ NonTerminationProof
              ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

G(a) → F(b)
H(X) → G(X)
F(X) → H(a)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

G(a) → F(b)
H(X) → G(X)
F(X) → H(a)

The TRS R consists of the following rules:none


s = F(X) evaluates to t =F(b)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

F(X) → H(a)
with rule F(X') → H(a) at position [] and matcher [X' / X]

H(a) → G(a)
with rule H(X) → G(X) at position [] and matcher [X / a]

G(a) → F(b)
with rule G(a) → F(b)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof

Q DP problem:
The TRS P consists of the following rules:

G(a) → F(b)
H(X) → G(X)
F(X) → H(a)

R is empty.
The set Q consists of the following terms:

h(x0)
g(a)
f(x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

h(x0)
g(a)
f(x0)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

G(a) → F(b)
H(X) → G(X)
F(X) → H(a)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.