Termination w.r.t. Q proof of /tmp/tmpp911Yf/minsort.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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(x, z)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(x, z)))

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:

MIN(x, cons(y, z)) → MIN(y, z)
MIN(x, cons(y, z)) → MIN(x, z)
MINSORT(cons(x, y)) → MIN(x, y)
MIN(x, cons(y, z)) → LE(x, y)
DEL(x, cons(y, z)) → EQ(x, y)
MINSORT(cons(x, y)) → DEL(min(x, y), cons(x, y))
EQ(s(x), s(y)) → EQ(x, y)
MIN(x, cons(y, z)) → IF(le(x, y), min(x, z), min(y, z))
MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))
DEL(x, cons(y, z)) → DEL(x, z)
DEL(x, cons(y, z)) → IF(eq(x, y), z, cons(y, del(x, z)))
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(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:

MIN(x, cons(y, z)) → MIN(y, z)
MIN(x, cons(y, z)) → MIN(x, z)
MINSORT(cons(x, y)) → MIN(x, y)
MIN(x, cons(y, z)) → LE(x, y)
DEL(x, cons(y, z)) → EQ(x, y)
MINSORT(cons(x, y)) → DEL(min(x, y), cons(x, y))
EQ(s(x), s(y)) → EQ(x, y)
MIN(x, cons(y, z)) → IF(le(x, y), min(x, z), min(y, z))
MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))
DEL(x, cons(y, z)) → DEL(x, z)
DEL(x, cons(y, z)) → IF(eq(x, y), z, cons(y, del(x, z)))
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(x, z)))

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.

EQ(s(x), s(y)) → EQ(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(EQ(x₁, x₂)) = x_2
POL(s(x₁)) = 1 + x_1

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(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:

DEL(x, cons(y, z)) → DEL(x, z)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(x, z)))

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.

DEL(x, cons(y, z)) → DEL(x, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 1/4 + (4)x_2
POL(DEL(x₁, x₂)) = (4)x_2

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(x, z)))

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.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x₁)) = 1 + x_1
POL(LE(x₁, x₂)) = x_2

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(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:

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

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(x, z)))

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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 4 + (4)x_2
POL(MIN(x₁, x₂)) = (4)x_2

The value of delta used in the strict ordering is 16.
The following usable rules [17] 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(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:

MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(x, z)))

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