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:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, 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:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, 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:

APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(eq, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y)))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(le, x), y)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(cons, app2(app2(min, x), y))
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(le, x)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(if, app2(app2(le, x), y))
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(del, x), z)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, y), z)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(eq, x), y)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(if, app2(app2(eq, x), y)), z)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, x), z)
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(cons, y), app2(app2(del, x), z))
APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(eq, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(min, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(min, x), y)
APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(app2(eq, x), y)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(del, app2(app2(min, x), y))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(le, x)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(if, app2(app2(eq, x), y))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z))
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(min, y)

The TRS R consists of the following rules:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, 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:

APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(eq, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y)))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(le, x), y)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(cons, app2(app2(min, x), y))
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(le, x)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(if, app2(app2(le, x), y))
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(del, x), z)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, y), z)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(eq, x), y)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(if, app2(app2(eq, x), y)), z)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, x), z)
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(cons, y), app2(app2(del, x), z))
APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(eq, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(min, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(min, x), y)
APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(app2(eq, x), y)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(del, app2(app2(min, x), y))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(le, x)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(if, app2(app2(eq, x), y))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z))
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(min, y)

The TRS R consists of the following rules:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 20 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(app2(eq, x), y)

The TRS R consists of the following rules:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(app2(eq, x), y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
eq  =  eq
s  =  s

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, 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
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(del, x), z)

The TRS R consists of the following rules:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(del, x), z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
del  =  del
cons  =  cons

Lexicographic Path Order [19].
Precedence:
[APP1, app1] > [del, cons]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, 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
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)

The TRS R consists of the following rules:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
le  =  le
s  =  s

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, 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
                ↳ QDPOrderProof
              ↳ QDP

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

APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, x), z)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, y), z)

The TRS R consists of the following rules:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, x), z)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, y), z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
min  =  min
cons  =  cons

Lexicographic Path Order [19].
Precedence:
APP1 > app1
APP1 > min
cons > app1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, 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:

APP2(minsort, app2(app2(cons, x), y)) -> APP2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y)))

The TRS R consists of the following rules:

app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))

The set Q consists of the following terms:

app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, x1), x2))

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