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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 QTRSRRRProof (⇔)
↳4 QTRS
↳5 DependencyPairsProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 QDP
↳9 MRRProof (⇔)
↳10 QDP
↳11 QDPOrderProof (⇔)
↳12 QDP
↳13 Instantiation (⇔)
↳14 QDP
↳15 NonTerminationProof (⇔)
↳16 FALSE

(0) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(head(x1)) = 1 + x1   
POL(incr(x1)) = x1   
POL(n__adx(x1)) = x1   
POL(n__incr(x1)) = x1   
POL(n__zeros) = 0   
POL(nats) = 1   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

nats → adx(zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)


(2) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(adx(x1)) = 1 + x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = x1   
POL(n__adx(x1)) = 1 + x1   
POL(n__incr(x1)) = x1   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

adx(nil) → nil


(4) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

INCR(cons(X, L)) → ACTIVATE(L)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ADX(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__adx(X)) → ADX(X)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

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 1 SCC with 1 less node.

(8) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ADX(cons(X, L)) → ACTIVATE(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

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

(9) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

ADX(cons(X, L)) → ACTIVATE(L)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(ADX(x1)) = 1 + x1   
POL(INCR(x1)) = x1   
POL(activate(x1)) = x1   
POL(adx(x1)) = 1 + x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = x1   
POL(n__adx(x1)) = 1 + x1   
POL(n__incr(x1)) = x1   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(zeros) = 0   

(10) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

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.


ACTIVATE(n__incr(X)) → INCR(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(ACTIVATE(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(n__incr(x1)) =
/0\
\1/
 +
/01\
\11/
·x1

POL(INCR(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/01\
\00/
·x1 +
/00\
\01/
·x2

POL(n__adx(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(ADX(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(activate(x1)) =
/1\
\0/
 +
/01\
\00/
·x1

POL(n__zeros) =
/1\
\0/

POL(zeros) =
/0\
\0/

POL(incr(x1)) =
/1\
\1/
 +
/01\
\00/
·x1

POL(nil) =
/0\
\0/

POL(0) =
/0\
\1/

POL(adx(x1)) =
/0\
\0/
 +
/11\
\11/
·x1

POL(s(x1)) =
/0\
\1/
 +
/00\
\11/
·x1

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

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

(13) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule INCR(cons(X, L)) → ACTIVATE(L) we obtained the following new rules [LPAR04]:

INCR(cons(y_0, n__adx(y_2))) → ACTIVATE(n__adx(y_2))

(14) Obligation:

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

ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
INCR(cons(y_0, n__adx(y_2))) → ACTIVATE(n__adx(y_2))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

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 = ADX(activate(n__zeros)) evaluates to t =ADX(activate(n__zeros))

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



Rewriting sequence

ADX(activate(n__zeros)) → ADX(zeros)
with rule activate(n__zeros) → zeros at position [0] and matcher [ ]

ADX(zeros) → ADX(cons(0, n__zeros))
with rule zeros → cons(0, n__zeros) at position [0] and matcher [ ]

ADX(cons(0, n__zeros)) → INCR(cons(0, n__adx(activate(n__zeros))))
with rule ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L)))) at position [] and matcher [X / 0, L / n__zeros]

INCR(cons(0, n__adx(activate(n__zeros)))) → ACTIVATE(n__adx(activate(n__zeros)))
with rule INCR(cons(y_0, n__adx(y_2))) → ACTIVATE(n__adx(y_2)) at position [] and matcher [y_0 / 0, y_2 / activate(n__zeros)]

ACTIVATE(n__adx(activate(n__zeros))) → ADX(activate(n__zeros))
with rule ACTIVATE(n__adx(X)) → ADX(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) FALSE