Termination w.r.t. Q proof of TRS/SK904.60.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

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:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(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:

MIN(x, .(y, z)) → MIN(y, z)
DEL(x, .(y, z)) → DEL(x, z)
MSORT(.(x, y)) → DEL(min(x, y), .(x, y))
MSORT(.(x, y)) → MIN(x, y)
MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))
MIN(x, .(y, z)) → MIN(x, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(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:

MIN(x, .(y, z)) → MIN(y, z)
DEL(x, .(y, z)) → DEL(x, z)
MSORT(.(x, y)) → DEL(min(x, y), .(x, y))
MSORT(.(x, y)) → MIN(x, y)
MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))
MIN(x, .(y, z)) → MIN(x, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(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:

MIN(x, .(y, z)) → MIN(y, z)
MSORT(.(x, y)) → MIN(x, y)
MSORT(.(x, y)) → DEL(min(x, y), .(x, y))
DEL(x, .(y, z)) → DEL(x, z)
MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))
MIN(x, .(y, z)) → MIN(x, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

DEL(x, .(y, z)) → DEL(x, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(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.

DEL(x, .(y, z)) → DEL(x, z)

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

Lexicographic path order with status [19].
Precedence:

trivial

Status:

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

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

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(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

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

MIN(x, .(y, z)) → MIN(y, z)
MIN(x, .(y, z)) → MIN(x, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(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.

MIN(x, .(y, z)) → MIN(y, z)
MIN(x, .(y, z)) → MIN(x, z)

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

Lexicographic path order with status [19].
Precedence:

trivial

Status:

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

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

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(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

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

MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(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.

MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MSORT(x1) = x1
.(x1, x2) = .
del(x1, x2) = del
if(x1, x2, x3) = if

Lexicographic path order with status [19].
Precedence:

. > del > if

Status:

trivial

The following usable rules [14] were oriented:

del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(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.