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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 QTRSRRRProof (⇔)
↳4 QTRS
↳5 QTRSRRRProof (⇔)
↳6 QTRS
↳7 DependencyPairsProof (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 QDP
↳11 UsableRulesReductionPairsProof (⇔)
↳12 QDP
↳13 NonTerminationProof (⇔)
↳14 NO

(0) Obligation:

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(c) → a__c
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a__c) = 0   
POL(a__f(x1, x2, x3)) = 1 + x1 + 2·x2 + x3   
POL(b) = 0   
POL(c) = 0   
POL(f(x1, x2, x3)) = 1 + x1 + 2·x2 + x3   
POL(mark(x1)) = 2·x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)


(2) Obligation:

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(c) → a__c
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a__c) = 0   
POL(a__f(x1, x2, x3)) = 1 + x1 + 2·x2 + x3   
POL(b) = 0   
POL(c) = 0   
POL(f(x1, x2, x3)) = 1 + x1 + 2·x2 + x3   
POL(mark(x1)) = 1 + 2·x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(c) → a__c
mark(b) → b


(4) Obligation:

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a__c) = 0   
POL(a__f(x1, x2, x3)) = 2 + x1 + 2·x2 + x3   
POL(b) = 0   
POL(c) = 0   
POL(f(x1, x2, x3)) = x1 + x2 + x3   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__f(X1, X2, X3) → f(X1, X2, X3)


(6) Obligation:

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__c → c

Q is empty.

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

A__F(b, X, c) → A__F(X, a__c, X)
A__F(b, X, c) → A__C

The TRS R consists of the following rules:

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__c → c

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Obligation:

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

A__F(b, X, c) → A__F(X, a__c, X)

The TRS R consists of the following rules:

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__c → c

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

(11) 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.

The following rules are removed from R:

a__f(b, X, c) → a__f(X, a__c, X)
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(A__F(x1, x2, x3)) = x1 + 2·x2 + x3   
POL(a__c) = 0   
POL(b) = 0   
POL(c) = 0   

(12) Obligation:

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

A__F(b, X, c) → A__F(X, a__c, X)

The TRS R consists of the following rules:

a__c → b
a__c → c

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

(13) 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 = A__F(a__c, X, a__c) evaluates to t =A__F(X, a__c, X)

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



Rewriting sequence

A__F(a__c, a__c, a__c) → A__F(a__c, a__c, c)
with rule a__c → c at position [2] and matcher [ ]

A__F(a__c, a__c, c) → A__F(b, a__c, c)
with rule a__c → b at position [0] and matcher [ ]

A__F(b, a__c, c) → A__F(a__c, a__c, a__c)
with rule A__F(b, X, c) → A__F(X, a__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.



(14) NO