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

0 QTRS
↳1 AAECC Innermost (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
      ↳26 Instantiation (⇔)
        ↳27 QDP
      ↳28 Instantiation (⇔)
        ↳29 QDP
        ↳30 Instantiation (⇔)
        ↳31 QDP
        ↳32 NonTerminationProof (⇔)
        ↳33 FALSE
  ↳34 QDP
    ↳35 UsableRulesProof (⇔)
    ↳36 QDP
    ↳37 QReductionProof (⇔)
    ↳38 QDP
    ↳39 QDPSizeChangeProof (⇔)
    ↳40 TRUE

(0) Obligation:

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

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

length(nil) → 0
length(cons(X, L)) → s(length(L))
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))

The TRS R 2 is

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false

The signature Sigma is {eq, true, false}

(2) Obligation:

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

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

EQ(s(X), s(Y)) → EQ(X, Y)
INF(X) → INF(s(X))
TAKE(s(X), cons(Y, L)) → TAKE(X, L)
LENGTH(cons(X, L)) → LENGTH(L)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs.

(6) Complex Obligation (AND)

(7) Obligation:

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

LENGTH(cons(X, L)) → LENGTH(L)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(8) 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.

(9) Obligation:

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

LENGTH(cons(X, L)) → LENGTH(L)

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

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(10) 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].

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

(11) Obligation:

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

LENGTH(cons(X, L)) → LENGTH(L)

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:

  • LENGTH(cons(X, L)) → LENGTH(L)
    The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

TAKE(s(X), cons(Y, L)) → TAKE(X, L)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(15) 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.

(16) Obligation:

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

TAKE(s(X), cons(Y, L)) → TAKE(X, L)

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

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(17) 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].

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

(18) Obligation:

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

TAKE(s(X), cons(Y, L)) → TAKE(X, L)

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

(19) 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:

  • TAKE(s(X), cons(Y, L)) → TAKE(X, L)
    The graph contains the following edges 1 > 1, 2 > 2

(20) TRUE

(21) Obligation:

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

INF(X) → INF(s(X))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(22) 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.

(23) Obligation:

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

INF(X) → INF(s(X))

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

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(24) 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].

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

(25) Obligation:

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

INF(X) → INF(s(X))

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

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule INF(X) → INF(s(X)) we obtained the following new rules [LPAR04]:

INF(s(z0)) → INF(s(s(z0)))

(27) Obligation:

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

INF(s(z0)) → INF(s(s(z0)))

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

(28) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule INF(X) → INF(s(X)) we obtained the following new rules [LPAR04]:

INF(s(z0)) → INF(s(s(z0)))

(29) Obligation:

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

INF(s(z0)) → INF(s(s(z0)))

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

(30) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule INF(s(z0)) → INF(s(s(z0))) we obtained the following new rules [LPAR04]:

INF(s(s(z0))) → INF(s(s(s(z0))))

(31) Obligation:

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

INF(s(s(z0))) → INF(s(s(s(z0))))

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

(32) 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 = INF(s(s(z0))) evaluates to t =INF(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [z0 / s(z0)]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from INF(s(s(z0))) to INF(s(s(s(z0)))).



(33) FALSE

(34) Obligation:

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

EQ(s(X), s(Y)) → EQ(X, Y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(35) 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.

(36) Obligation:

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

EQ(s(X), s(Y)) → EQ(X, Y)

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

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(37) 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].

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

(38) Obligation:

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

EQ(s(X), s(Y)) → EQ(X, Y)

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

(39) 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:

  • EQ(s(X), s(Y)) → EQ(X, Y)
    The graph contains the following edges 1 > 1, 2 > 2

(40) TRUE