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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 YES
    ↳14 QDP
        ↳15 NonTerminationProof (⇔)
        ↳16 NO

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

A__F(a, b, X) → A__F(mark(X), X, mark(X))
A__F(a, b, X) → MARK(X)
MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3))
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(X1, X2, X3)) → MARK(X3)
MARK(c) → A__C

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

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

(4) Obligation:

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

A__F(a, b, X) → MARK(X)
MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3))
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(X1, X2, X3)) → MARK(X3)

The TRS R consists of the following rules:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3))
MARK(f(X1, X2, X3)) → MARK(X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(A__F(x1, x2, x3)) = -I + -I·x1 + -I·x2 + 0A·x3

POL(a) = 0A

POL(b) = 0A

POL(MARK(x1)) = -I + 0A·x1

POL(f(x1, x2, x3)) = -I + 0A·x1 + 0A·x2 + 1A·x3

POL(mark(x1)) = -I + 0A·x1

POL(a__f(x1, x2, x3)) = -I + 0A·x1 + 0A·x2 + 1A·x3

POL(c) = 0A

POL(a__c) = 0A

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

A__F(a, b, X) → MARK(X)
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

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

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Complex Obligation (AND)

(9) Obligation:

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

MARK(f(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

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

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

(10) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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

MARK(f(X1, X2, X3)) → MARK(X1)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MARK(f(X1, X2, X3)) → MARK(X1)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

A__F(a, b, X) → A__F(mark(X), X, mark(X))

The TRS R consists of the following rules:

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

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

(15) 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(mark(mark(a__c)), mark(a__c), mark(mark(a__c))) evaluates to t =A__F(mark(mark(a__c)), mark(a__c), mark(mark(a__c)))

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



Rewriting sequence

A__F(mark(mark(a__c)), mark(a__c), mark(mark(a__c))) → A__F(mark(mark(a__c)), mark(a__c), mark(mark(c)))
with rule a__c → c at position [2,0,0] and matcher [ ]

A__F(mark(mark(a__c)), mark(a__c), mark(mark(c))) → A__F(mark(mark(a__c)), mark(a__c), mark(a__c))
with rule mark(c) → a__c at position [2,0] and matcher [ ]

A__F(mark(mark(a__c)), mark(a__c), mark(a__c)) → A__F(mark(mark(a__c)), mark(a__c), mark(c))
with rule a__c → c at position [2,0] and matcher [ ]

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

A__F(mark(mark(a__c)), mark(a__c), a__c) → A__F(mark(mark(a__c)), mark(b), a__c)
with rule a__c → b at position [1,0] and matcher [ ]

A__F(mark(mark(a__c)), mark(b), a__c) → A__F(mark(mark(a__c)), b, a__c)
with rule mark(b) → b at position [1] and matcher [ ]

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

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

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

A__F(a, b, a__c) → A__F(mark(a__c), a__c, mark(a__c))
with rule A__F(a, b, X') → A__F(mark(X'), X', mark(X')) at position [] and matcher [X' / a__c]

A__F(mark(a__c), a__c, mark(a__c)) → A__F(mark(a__c), b, mark(a__c))
with rule a__c → b at position [1] and matcher [ ]

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

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

A__F(a, b, mark(a__c)) → A__F(mark(mark(a__c)), mark(a__c), mark(mark(a__c)))
with rule A__F(a, b, X) → A__F(mark(X), X, mark(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.



(16) NO