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

0 QTRS
↳1 QTRSRRRProof (⇔, 86 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 QDPSizeChangeProof (⇔, 0 ms)
        ↳11 YES
    ↳12 QDP
        ↳13 NonLoopProof (⇔, 41 ms)
        ↳14 NO
    ↳15 QDP
        ↳16 NonLoopProof (⇔, 0 ms)
        ↳17 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) 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

(11) YES

(12) 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.

(13) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [ ] on the rule
PAIRS[ ]n[ ] → PAIRS[ ]n[ ]
This rule is correct for the QDP as the following derivation shows:

PAIRS[ ]n[ ] → PAIRS[ ]n[ ]
    by Narrowing at position: []
        PAIRS[ ]n[ ] → ODDS[ ]n[ ]
            by OriginalRule from TRS P

        ODDS[ ]n[ ] → PAIRS[ ]n[ ]
            by OriginalRule from TRS P

(14) NO

(15) 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.

(16) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [ ] on the rule
NATS[ ]n[ ] → NATS[ ]n[ ]
This rule is correct for the QDP as the following derivation shows:

NATS[ ]n[ ] → NATS[ ]n[ ]
    by OriginalRule from TRS P

(17) NO