NO Termination w.r.t. Q proof of Transformed_CSR_04_Ex1_Luc04b_FR.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 9 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 TransformationProof (⇔, 0 ms)
↳10 QDP
↳11 TransformationProof (⇔, 0 ms)
↳12 QDP
↳13 TransformationProof (⇔, 0 ms)
↳14 QDP
↳15 DependencyGraphProof (⇔, 0 ms)
↳16 QDP
↳17 NonTerminationLoopProof (⇒, 0 ms)
↳18 NO

(0) Obligation:

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

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(head(x1)) = 2 + x1   
POL(incr(x1)) = 2·x1   
POL(n__incr(x1)) = 2·x1   
POL(n__nats) = 0   
POL(n__odds) = 0   
POL(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 2·x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

head(cons(X, XS)) → X


(2) Obligation:

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

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(incr(x1)) = 2·x1   
POL(n__incr(x1)) = 2·x1   
POL(n__nats) = 0   
POL(n__odds) = 0   
POL(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = 1 + 2·x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

tail(cons(X, XS)) → activate(XS)


(4) Obligation:

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

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
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:

ODDS → INCR(pairs)
ODDS → PAIRS
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__nats) → NATS
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
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 2 less nodes.

(8) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__odds) → ODDS
ODDS → INCR(pairs)

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

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

(9) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ODDS → INCR(pairs) at position [0] we obtained the following new rules [LPAR04]:

ODDS → INCR(cons(0, n__incr(n__odds))) → ODDS → INCR(cons(0, n__incr(n__odds)))

(10) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__odds) → ODDS
ODDS → INCR(cons(0, n__incr(n__odds)))

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

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

(11) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__incr(X)) → INCR(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0))) → ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__nats)) → INCR(nats) → ACTIVATE(n__incr(n__nats)) → INCR(nats)
ACTIVATE(n__incr(n__odds)) → INCR(odds) → ACTIVATE(n__incr(n__odds)) → INCR(odds)
ACTIVATE(n__incr(x0)) → INCR(x0) → ACTIVATE(n__incr(x0)) → INCR(x0)

(12) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__odds) → ODDS
ODDS → INCR(cons(0, n__incr(n__odds)))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__nats)) → INCR(nats)
ACTIVATE(n__incr(n__odds)) → INCR(odds)
ACTIVATE(n__incr(x0)) → INCR(x0)

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

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

(13) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__incr(n__nats)) → INCR(nats) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats))) → ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats)))
ACTIVATE(n__incr(n__nats)) → INCR(n__nats) → ACTIVATE(n__incr(n__nats)) → INCR(n__nats)

(14) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__odds) → ODDS
ODDS → INCR(cons(0, n__incr(n__odds)))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__odds)) → INCR(odds)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats)))
ACTIVATE(n__incr(n__nats)) → INCR(n__nats)

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Obligation:

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

ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__odds) → ODDS
ODDS → INCR(cons(0, n__incr(n__odds)))
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__odds)) → INCR(odds)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats)))

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

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

(17) NonTerminationLoopProof (COMPLETE 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 = ACTIVATE(n__incr(n__nats)) evaluates to t =ACTIVATE(n__incr(n__nats))

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



Rewriting sequence

ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats)))
with rule ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats))) at position [] and matcher [ ]

INCR(cons(0, n__incr(n__nats))) → ACTIVATE(n__incr(n__nats))
with rule INCR(cons(X, XS)) → ACTIVATE(XS)

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.



(18) NO