Termination w.r.t. Q proof of /tmp/tmpXdhuQ-/thiemann08.trs

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:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IF(false, x, l) → TAIL(l)
IF(false, x, l) → HEAD(l)
REV(cons(x, l)) → REV2(x, l)
IF(false, x, l) → LAST(head(l), tail(l))
REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
LAST(x, l) → EMPTY(l)
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

IF(false, x, l) → TAIL(l)
IF(false, x, l) → HEAD(l)
REV(cons(x, l)) → REV2(x, l)
IF(false, x, l) → LAST(head(l), tail(l))
REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
LAST(x, l) → EMPTY(l)
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

IF(false, x, l) → LAST(head(l), tail(l))
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The remaining pairs can at least be oriented weakly.

REV(cons(x, l)) → REV2(x, l)

Used ordering: Polynomial interpretation [25,35]:

POL(REV2(x₁, x₂)) = 2 + x_2
POL(cons(x₁, x₂)) = 1 + x_2
POL(rev2(x₁, x₂)) = x_2
POL(REV(x₁)) = 1 + x_1
POL(nil) = 0
POL(rev(x₁)) = x_1
POL(rev1(x₁, x₂)) = (3)x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

REV(cons(x, l)) → REV2(x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.