Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/Ste92SLASHminsort.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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(x1, x2))

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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(x1, x2))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(x1, x2))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

Used argument filtering: EQ₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DEL₂(x, cons₂(y, z)) -> DEL₂(x, z)

Used argument filtering: DEL₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

Used argument filtering: LE₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: MIN₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(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

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
minsort₁(nil)
minsort₁(cons₂(x0, x1))
min₂(x0, nil)
min₂(x0, cons₂(x1, x2))
del₂(x0, nil)
del₂(x0, cons₂(x1, x2))

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