Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 UsableRulesReductionPairsProof (⇔)
↳8 QDP
↳9 NonTerminationProof (⇔)
↳10 FALSE

(0) Obligation:

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

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c
activate(n__b) → b
activate(n__c) → c
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(activate(x1)) = 2 + 2·x1   
POL(b) = 1   
POL(c) = 1   
POL(f(x1, x2, x3)) = x1 + 2·x2 + x3   
POL(n__b) = 1   
POL(n__c) = 1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

activate(n__b) → b
activate(n__c) → c
activate(X) → X


(2) Obligation:

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

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

F(n__b, X, n__c) → F(X, c, X)
F(n__b, X, n__c) → C
C → B

The TRS R consists of the following rules:

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(6) Obligation:

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

F(n__b, X, n__c) → F(X, c, X)

The TRS R consists of the following rules:

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c

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

(7) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F(x1, x2, x3)) = x1 + 2·x2 + x3   
POL(b) = 0   
POL(c) = 0   
POL(n__b) = 0   
POL(n__c) = 0   

(8) Obligation:

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

F(n__b, X, n__c) → F(X, c, X)

The TRS R consists of the following rules:

c → b
c → n__c
b → n__b

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

(9) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = F(c, X, c) evaluates to t =F(X, c, X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [X / c]
  • Matcher: [ ]



Rewriting sequence

F(c, c, c) → F(c, c, n__c)
with rule c → n__c at position [2] and matcher [ ]

F(c, c, n__c) → F(b, c, n__c)
with rule c → b at position [0] and matcher [ ]

F(b, c, n__c) → F(n__b, c, n__c)
with rule b → n__b at position [0] and matcher [ ]

F(n__b, c, n__c) → F(c, c, c)
with rule F(n__b, X, n__c) → F(X, c, X)

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.



(10) FALSE