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 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔)
        ↳16 QDP
        ↳17 NonTerminationProof (⇔)
        ↳18 NO
    ↳19 QDP
        ↳20 UsableRulesProof (⇔)
        ↳21 QDP
        ↳22 NonTerminationProof (⇔)
        ↳23 NO

(0) Obligation:

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

nats → cons(0, n__incr(nats))
pairs → cons(0, n__incr(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)
activate(n__incr(X)) → incr(X)
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)) = 2·x1 + 2·x2   
POL(head(x1)) = 2 + 2·x1   
POL(incr(x1)) = x1   
POL(n__incr(x1)) = x1   
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(nats))
pairs → cons(0, n__incr(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)
activate(n__incr(X)) → incr(X)
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(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = 2 + 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(nats))
pairs → cons(0, n__incr(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
activate(n__incr(X)) → incr(X)
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:

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

The TRS R consists of the following rules:

nats → cons(0, n__incr(nats))
pairs → cons(0, n__incr(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
activate(n__incr(X)) → incr(X)
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 3 SCCs with 1 less node.

(8) Complex Obligation (AND)

(9) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

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

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:

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, XS)) → ACTIVATE(XS)

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:

  • INCR(cons(X, XS)) → ACTIVATE(XS)
    The graph contains the following edges 1 > 1

  • ACTIVATE(n__incr(X)) → INCR(X)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

ODDS → PAIRS
PAIRS → ODDS

The TRS R consists of the following rules:

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

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

(15) 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.

(16) Obligation:

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

ODDS → PAIRS
PAIRS → ODDS

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

(17) 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 = PAIRS evaluates to t =PAIRS

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



Rewriting sequence

PAIRS → ODDS
with rule PAIRS → ODDS at position [] and matcher [ ]

ODDS → PAIRS
with rule ODDS → PAIRS

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

(19) Obligation:

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

NATS → NATS

The TRS R consists of the following rules:

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

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

(20) 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.

(21) Obligation:

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

NATS → NATS

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

(22) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = NATS evaluates to t =NATS

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from NATS to NATS.



(23) NO