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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 QTRSRRRProof (⇔)
↳4 QTRS
↳5 QTRSRRRProof (⇔)
↳6 QTRS
↳7 QTRSRRRProof (⇔)
↳8 QTRS
↳9 QTRSRRRProof (⇔)
↳10 QTRS
↳11 Overlay + Local Confluence (⇔)
↳12 QTRS
↳13 DependencyPairsProof (⇔)
↳14 QDP
↳15 DependencyGraphProof (⇔)
↳16 AND
  ↳17 QDP
    ↳18 UsableRulesProof (⇔)
    ↳19 QDP
    ↳20 QReductionProof (⇔)
    ↳21 QDP
    ↳22 QDPSizeChangeProof (⇔)
    ↳23 TRUE
  ↳24 QDP
    ↳25 UsableRulesProof (⇔)
    ↳26 QDP
    ↳27 QReductionProof (⇔)
    ↳28 QDP
    ↳29 MRRProof (⇔)
    ↳30 QDP
    ↳31 Narrowing (⇔)
    ↳32 QDP
    ↳33 UsableRulesProof (⇔)
    ↳34 QDP
    ↳35 QReductionProof (⇔)
    ↳36 QDP
    ↳37 Rewriting (⇔)
    ↳38 QDP
    ↳39 UsableRulesProof (⇔)
    ↳40 QDP
    ↳41 QReductionProof (⇔)
    ↳42 QDP
    ↳43 Instantiation (⇔)
    ↳44 QDP
    ↳45 NonTerminationProof (⇔)
    ↳46 FALSE

(0) Obligation:

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

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__and(x1, x2)) = x1 + x2   
POL(a__length(x1)) = x1   
POL(a__zeros) = 1   
POL(and(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(length(x1)) = x1   
POL(mark(x1)) = 1 + x1   
POL(nil) = 1   
POL(s(x1)) = x1   
POL(tt) = 2   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__and(tt, X) → mark(X)
a__length(nil) → 0
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
a__zeros → zeros


(2) Obligation:

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__and(x1, x2)) = x1 + 2·x2   
POL(a__length(x1)) = 1 + x1   
POL(a__zeros) = 0   
POL(and(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(length(x1)) = 1 + x1   
POL(mark(x1)) = 2·x1   
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:

mark(length(X)) → a__length(mark(X))


(4) Obligation:

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__and(x1, x2)) = 1 + x1 + 2·x2   
POL(a__length(x1)) = 2·x1   
POL(a__zeros) = 0   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(length(x1)) = x1   
POL(mark(x1)) = 2·x1   
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:

mark(and(X1, X2)) → a__and(mark(X1), X2)


(6) Obligation:

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__and(x1, x2)) = 1 + x1 + x2   
POL(a__length(x1)) = 1 + x1   
POL(a__zeros) = 0   
POL(and(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(length(x1)) = x1   
POL(mark(x1)) = x1   
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:

a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)


(8) Obligation:

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))

Q is empty.

(9) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__length(x1)) = 2·x1   
POL(a__zeros) = 2   
POL(cons(x1, x2)) = 2 + x1 + 2·x2   
POL(mark(x1)) = 2 + 2·x1   
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:

mark(cons(X1, X2)) → cons(mark(X1), X2)


(10) Obligation:

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

Q is empty.

(11) Overlay + Local Confluence (EQUIVALENT transformation)

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

(12) Obligation:

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

(13) DependencyPairsProof (EQUIVALENT transformation)

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

(14) Obligation:

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

A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
A__LENGTH(cons(N, L)) → MARK(L)
MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.

(16) Complex Obligation (AND)

(17) Obligation:

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

MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

(18) 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.

(19) Obligation:

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

MARK(s(X)) → MARK(X)

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

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

(20) 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].

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

(21) Obligation:

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

MARK(s(X)) → MARK(X)

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

(22) 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:

  • MARK(s(X)) → MARK(X)
    The graph contains the following edges 1 > 1

(23) TRUE

(24) Obligation:

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

A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

(25) 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.

(26) Obligation:

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

A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

(27) 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].

a__length(cons(x0, x1))

(28) Obligation:

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

A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
mark(zeros)
mark(s(x0))

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

(29) 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 rules of the TRS R:

mark(s(X)) → s(mark(X))

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__LENGTH(x1)) = 2·x1   
POL(a__zeros) = 0   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(mark(x1)) = 2·x1   
POL(s(x1)) = 1 + 2·x1   
POL(zeros) = 0   

(30) Obligation:

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

A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

mark(zeros) → a__zeros
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
mark(zeros)
mark(s(x0))

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

(31) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule A__LENGTH(cons(N, L)) → A__LENGTH(mark(L)) at position [0] we obtained the following new rules [LPAR04]:

A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)

(32) Obligation:

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)

The TRS R consists of the following rules:

mark(zeros) → a__zeros
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
mark(zeros)
mark(s(x0))

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

(33) 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.

(34) Obligation:

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
mark(zeros)
mark(s(x0))

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

(35) 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].

mark(zeros)
mark(s(x0))

(36) Obligation:

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros

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

(37) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros) at position [0] we obtained the following new rules [LPAR04]:

A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros))

(38) Obligation:

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros

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

(39) 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.

(40) Obligation:

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros))

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

a__zeros

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

(41) 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].

a__zeros

(42) Obligation:

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros))

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

(43) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros)) we obtained the following new rules [LPAR04]:

A__LENGTH(cons(0, zeros)) → A__LENGTH(cons(0, zeros))

(44) Obligation:

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

A__LENGTH(cons(0, zeros)) → A__LENGTH(cons(0, zeros))

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

(45) 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 = A__LENGTH(cons(0, zeros)) evaluates to t =A__LENGTH(cons(0, 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 A__LENGTH(cons(0, zeros)) to A__LENGTH(cons(0, zeros)).



(46) FALSE