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 Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 3


POL( EQ2(x1, x2) ) = 3x2 + 3



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 Order [17,21] with Interpretation:

POL( cons2(x1, x2) ) = 3x1 + x2 + 1


POL( DEL2(x1, x2) ) = x1 + 2x2 + 3



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 Order [17,21] with Interpretation:

POL( LE2(x1, x2) ) = 3x2 + 3


POL( s1(x1) ) = x1 + 3



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 Order [17,21] with Interpretation:

POL( MIN2(x1, x2) ) = 3x1 + x2 + 1


POL( cons2(x1, x2) ) = 3x1 + 3x2 + 3



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.