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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 Overlay + Local Confluence (⇔)
↳4 QTRS
↳5 DependencyPairsProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔)
        ↳11 QDP
        ↳12 QReductionProof (⇔)
        ↳13 QDP
        ↳14 QDPSizeChangeProof (⇔)
        ↳15 YES
    ↳16 QDP
        ↳17 UsableRulesProof (⇔)
        ↳18 QDP
        ↳19 QReductionProof (⇔)
        ↳20 QDP
        ↳21 QDPSizeChangeProof (⇔)
        ↳22 YES
    ↳23 QDP
        ↳24 UsableRulesProof (⇔)
        ↳25 QDP
        ↳26 QReductionProof (⇔)
        ↳27 QDP
        ↳28 NonTerminationProof (⇔)
        ↳29 NO

(0) Obligation:

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

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(hd(x1)) = 2 + 2·x1   
POL(incr(x1)) = x1   
POL(nats) = 2   
POL(s(x1)) = x1   
POL(tl(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)
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y


(2) Obligation:

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

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(4) Obligation:

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

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

(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:

ZEROS → ZEROS
INCR(cons(X, Y)) → INCR(Y)
ADX(cons(X, Y)) → INCR(cons(X, adx(Y)))
ADX(cons(X, Y)) → ADX(Y)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

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:

INCR(cons(X, Y)) → INCR(Y)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

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

(10) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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

INCR(cons(X, Y)) → INCR(Y)

R is empty.
The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

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

(12) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

(13) Obligation:

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

INCR(cons(X, Y)) → INCR(Y)

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

(14) 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, Y)) → INCR(Y)
    The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

ADX(cons(X, Y)) → ADX(Y)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

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

(17) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(18) Obligation:

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

ADX(cons(X, Y)) → ADX(Y)

R is empty.
The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

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

(19) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

(20) Obligation:

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

ADX(cons(X, Y)) → ADX(Y)

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

(21) 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:

  • ADX(cons(X, Y)) → ADX(Y)
    The graph contains the following edges 1 > 1

(22) YES

(23) Obligation:

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

ZEROS → ZEROS

The TRS R consists of the following rules:

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

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

(24) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(25) Obligation:

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

ZEROS → ZEROS

R is empty.
The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

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

(26) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

(27) Obligation:

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

ZEROS → ZEROS

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

(28) 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 = ZEROS evaluates to t =ZEROS

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 ZEROS to ZEROS.



(29) NO