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:

g(X) → h(X)
c → d
h(d) → g(c)

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

g(X) → h(X)
c → d
h(d) → g(c)

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

c → d

The TRS R 2 is

g(X) → h(X)
h(d) → g(c)

The signature Sigma is {g, h}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

g(X) → h(X)
c → d
h(d) → g(c)

The set Q consists of the following terms:

g(x0)
c
h(d)


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

H(d) → C
H(d) → G(c)
G(X) → H(X)

The TRS R consists of the following rules:

g(X) → h(X)
c → d
h(d) → g(c)

The set Q consists of the following terms:

g(x0)
c
h(d)

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

H(d) → C
H(d) → G(c)
G(X) → H(X)

The TRS R consists of the following rules:

g(X) → h(X)
c → d
h(d) → g(c)

The set Q consists of the following terms:

g(x0)
c
h(d)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
              ↳ UsableRulesProof

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

H(d) → G(c)
G(X) → H(X)

The TRS R consists of the following rules:

g(X) → h(X)
c → d
h(d) → g(c)

The set Q consists of the following terms:

g(x0)
c
h(d)

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

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

H(d) → G(c)
G(X) → H(X)

The TRS R consists of the following rules:

c → d

The set Q consists of the following terms:

g(x0)
c
h(d)

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.

g(x0)
h(d)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ MNOCProof
              ↳ UsableRulesProof

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

H(d) → G(c)
G(X) → H(X)

The TRS R consists of the following rules:

c → d

The set Q consists of the following terms:

c

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ MNOCProof
                        ↳ QDP
                          ↳ NonTerminationProof
              ↳ UsableRulesProof

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

H(d) → G(c)
G(X) → H(X)

The TRS R consists of the following rules:

c → d

Q is empty.
We have to consider all (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:

H(d) → G(c)
G(X) → H(X)

The TRS R consists of the following rules:

c → d


s = G(c) evaluates to t =G(c)

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



Rewriting sequence

G(c) → G(d)
with rule c → d at position [0] and matcher [ ]

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

H(d) → G(c)
with rule H(d) → G(c)

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

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

H(d) → G(c)
G(X) → H(X)

The TRS R consists of the following rules:

c → d

The set Q consists of the following terms:

g(x0)
c
h(d)

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.

g(x0)
h(d)



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

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

H(d) → G(c)
G(X) → H(X)

The TRS R consists of the following rules:

c → d

The set Q consists of the following terms:

c

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