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(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
c → n__c
activate(n__d) → d
activate(n__c) → c
activate(X) → X

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
c → n__c
activate(n__d) → d
activate(n__c) → c
activate(X) → X

Q is empty.

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

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
c → n__c
activate(n__d) → d
activate(n__c) → c
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:

c → n__c
activate(n__d) → d
activate(X) → X
Used ordering:
Polynomial interpretation [25]:

POL(activate(x1)) = 1 + 2·x1   
POL(c) = 1   
POL(d) = 1   
POL(g(x1)) = 1 + 2·x1   
POL(h(x1)) = x1   
POL(n__c) = 0   
POL(n__d) = 1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost

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

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c

Q is empty.

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

d → n__d
activate(n__c) → c
c → d

The TRS R 2 is

g(X) → h(activate(X))
h(n__d) → g(n__c)

The signature Sigma is {g, h}

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

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

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)


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(n__d) → G(n__c)
C → D
ACTIVATE(n__c) → C
G(X) → H(activate(X))
G(X) → ACTIVATE(X)

The TRS R consists of the following rules:

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)

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:

H(n__d) → G(n__c)
C → D
ACTIVATE(n__c) → C
G(X) → H(activate(X))
G(X) → ACTIVATE(X)

The TRS R consists of the following rules:

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

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

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

H(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)

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
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof
                  ↳ UsableRulesProof

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

H(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

activate(n__c) → c
c → d
d → n__d

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)

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(n__d)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ 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(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

activate(n__c) → c
c → d
d → n__d

The set Q consists of the following terms:

c
d
activate(n__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
  ↳ RRRPoloQTRSProof
    ↳ 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(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

activate(n__c) → c
c → d
d → n__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(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

activate(n__c) → c
c → d
d → n__d


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

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



Rewriting sequence

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

H(activate(n__c)) → H(c)
with rule activate(n__c) → c at position [0] and matcher [ ]

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

H(d) → H(n__d)
with rule d → n__d at position [0] and matcher [ ]

H(n__d) → G(n__c)
with rule H(n__d) → G(n__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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ UsableRulesProof
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof

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

H(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

activate(n__c) → c
c → d
d → n__d

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)

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(n__d)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ 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(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

activate(n__c) → c
c → d
d → n__d

The set Q consists of the following terms:

c
d
activate(n__c)

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