Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.


↳ QTRS
  ↳ AAECC Innermost

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.

We have applied [15,7] 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 {true, false, eq}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

EQ(s(X), s(Y)) → EQ(X, Y)
TAKE(s(X), cons(Y, L)) → TAKE(X, L)
LENGTH(cons(X, L)) → LENGTH(L)
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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

EQ(s(X), s(Y)) → EQ(X, Y)
TAKE(s(X), cons(Y, L)) → TAKE(X, L)
LENGTH(cons(X, L)) → LENGTH(L)
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.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LENGTH(cons(X, L)) → LENGTH(L)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LENGTH(x1)  =  x1
cons(x1, x2)  =  cons(x2)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

Q DP problem:
P is empty.
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TAKE(s(X), cons(Y, L)) → TAKE(X, L)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

Q DP problem:
P is empty.
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


EQ(s(X), s(Y)) → EQ(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

Q DP problem:
P is empty.
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.