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:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)


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

REV(xs, ys) → DROPLAST(xs)
REV(xs, ys) → ISEMPTY(xs)
REV(xs, ys) → LAST(xs)
REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))
REVERSE(xs) → REV(xs, nil)
IF(false, xs, ys, zs) → REV(xs, ys)
APPEND(cons(x, xs), ys) → APPEND(xs, ys)
REV(xs, ys) → APPEND(ys, last(xs))
LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

REV(xs, ys) → DROPLAST(xs)
REV(xs, ys) → ISEMPTY(xs)
REV(xs, ys) → LAST(xs)
REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))
REVERSE(xs) → REV(xs, nil)
IF(false, xs, ys, zs) → REV(xs, ys)
APPEND(cons(x, xs), ys) → APPEND(xs, ys)
REV(xs, ys) → APPEND(ys, last(xs))
LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

REV(xs, ys) → ISEMPTY(xs)
REV(xs, ys) → DROPLAST(xs)
REV(xs, ys) → LAST(xs)
REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))
REVERSE(xs) → REV(xs, nil)
APPEND(cons(x, xs), ys) → APPEND(xs, ys)
IF(false, xs, ys, zs) → REV(xs, ys)
LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))
REV(xs, ys) → APPEND(ys, last(xs))

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 5 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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.


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

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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.


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

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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.


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

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
IF(false, xs, ys, zs) → REV(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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