Termination w.r.t. Q proof of TRS/Ste92minsort.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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

Q is empty.

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

DEL₂(x, cons₂(y, z)) -> IF₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))
MIN₂(x, cons₂(y, z)) -> MIN₂(x, z)
EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
DEL₂(x, cons₂(y, z)) -> DEL₂(x, z)
MINSORT₁(cons₂(x, y)) -> DEL₂(min₂(x, y), cons₂(x, y))
MIN₂(x, cons₂(y, z)) -> MIN₂(y, z)
MIN₂(x, cons₂(y, z)) -> IF₃(le₂(x, y), min₂(x, z), min₂(y, z))
MINSORT₁(cons₂(x, y)) -> MIN₂(x, y)
MIN₂(x, cons₂(y, z)) -> LE₂(x, y)
MINSORT₁(cons₂(x, y)) -> MINSORT₁(del₂(min₂(x, y), cons₂(x, y)))
DEL₂(x, cons₂(y, z)) -> EQ₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

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:

DEL₂(x, cons₂(y, z)) -> IF₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))
MIN₂(x, cons₂(y, z)) -> MIN₂(x, z)
EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
DEL₂(x, cons₂(y, z)) -> DEL₂(x, z)
MINSORT₁(cons₂(x, y)) -> DEL₂(min₂(x, y), cons₂(x, y))
MIN₂(x, cons₂(y, z)) -> MIN₂(y, z)
MIN₂(x, cons₂(y, z)) -> IF₃(le₂(x, y), min₂(x, z), min₂(y, z))
MINSORT₁(cons₂(x, y)) -> MIN₂(x, y)
MIN₂(x, cons₂(y, z)) -> LE₂(x, y)
MINSORT₁(cons₂(x, y)) -> MINSORT₁(del₂(min₂(x, y), cons₂(x, y)))
DEL₂(x, cons₂(y, z)) -> EQ₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

Q is empty.
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.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(EQ₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DEL₂(x, cons₂(y, z)) -> DEL₂(x, z)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

Q is empty.
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.

DEL₂(x, cons₂(y, z)) -> DEL₂(x, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DEL₂(x₁, x₂)) = x₂
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

Q is empty.
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.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

MIN₂(x, cons₂(y, z)) -> MIN₂(x, z)
MIN₂(x, cons₂(y, z)) -> MIN₂(y, z)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

Q is empty.
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.

MIN₂(x, cons₂(y, z)) -> MIN₂(x, z)
MIN₂(x, cons₂(y, z)) -> MIN₂(y, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MIN₂(x₁, x₂)) = x₂
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINSORT₁(cons₂(x, y)) -> MINSORT₁(del₂(min₂(x, y), cons₂(x, y)))

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
eq₂(0, 0) -> true
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
minsort₁(nil) -> nil
minsort₁(cons₂(x, y)) -> cons₂(min₂(x, y), minsort₁(del₂(min₂(x, y), cons₂(x, y))))
min₂(x, nil) -> x
min₂(x, cons₂(y, z)) -> if₃(le₂(x, y), min₂(x, z), min₂(y, z))
del₂(x, nil) -> nil
del₂(x, cons₂(y, z)) -> if₃(eq₂(x, y), z, cons₂(y, del₂(x, z)))

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