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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 NonTerminationProof (⇔)
    ↳14 FALSE

(0) Obligation:

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

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

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:

ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
ACTIVE(f(a, b, X)) → F(X, X, X)
ACTIVE(c) → MARK(a)
ACTIVE(c) → MARK(b)
MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, mark(X3)))
MARK(f(X1, X2, X3)) → F(X1, X2, mark(X3))
MARK(f(X1, X2, X3)) → MARK(X3)
MARK(a) → ACTIVE(a)
MARK(b) → ACTIVE(b)
MARK(c) → ACTIVE(c)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

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

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 2 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(X1, mark(X2), X3) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

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

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

(6) 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.

(7) Obligation:

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

F(X1, mark(X2), X3) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

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

(8) 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:

  • F(X1, mark(X2), X3) → F(X1, X2, X3)
    The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3

  • F(mark(X1), X2, X3) → F(X1, X2, X3)
    The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3

  • F(X1, X2, mark(X3)) → F(X1, X2, X3)
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

  • F(active(X1), X2, X3) → F(X1, X2, X3)
    The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3

  • F(X1, active(X2), X3) → F(X1, X2, X3)
    The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3

  • F(X1, X2, active(X3)) → F(X1, X2, X3)
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(11) 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)) → MARK(X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(a) = 0   
POL(active(x1)) = x1   
POL(b) = 0   
POL(c) = 0   
POL(f(x1, x2, x3)) = 1 + x3   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

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

The TRS R consists of the following rules:

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

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 = ACTIVE(f(active(c), active(c), mark(X3))) evaluates to t =ACTIVE(f(X3, X3, mark(X3)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [X3 / active(c)]
  • Matcher: [ ]



Rewriting sequence

ACTIVE(f(active(c), active(c), mark(active(c)))) → ACTIVE(f(active(c), mark(b), mark(active(c))))
with rule active(c) → mark(b) at position [0,1] and matcher [ ]

ACTIVE(f(active(c), mark(b), mark(active(c)))) → ACTIVE(f(mark(a), mark(b), mark(active(c))))
with rule active(c) → mark(a) at position [0,0] and matcher [ ]

ACTIVE(f(mark(a), mark(b), mark(active(c)))) → ACTIVE(f(mark(a), mark(b), active(c)))
with rule f(X1, X2, mark(X3)) → f(X1, X2, X3) at position [0] and matcher [X1 / mark(a), X2 / mark(b), X3 / active(c)]

ACTIVE(f(mark(a), mark(b), active(c))) → ACTIVE(f(mark(a), b, active(c)))
with rule f(X1, mark(X2), X3') → f(X1, X2, X3') at position [0] and matcher [X1 / mark(a), X2 / b, X3' / active(c)]

ACTIVE(f(mark(a), b, active(c))) → ACTIVE(f(a, b, active(c)))
with rule f(mark(X1), X2, X3) → f(X1, X2, X3) at position [0] and matcher [X1 / a, X2 / b, X3 / active(c)]

ACTIVE(f(a, b, active(c))) → MARK(f(active(c), active(c), active(c)))
with rule ACTIVE(f(a, b, X)) → MARK(f(X, X, X)) at position [] and matcher [X / active(c)]

MARK(f(active(c), active(c), active(c))) → ACTIVE(f(active(c), active(c), mark(active(c))))
with rule MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, mark(X3)))

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) FALSE