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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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:

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(del, x), app(app(cons, y), z)) → APP(eq, x)
APP(minsort, app(app(cons, x), y)) → APP(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(min, x), app(app(cons, y), z)) → APP(app(le, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(minsort, app(app(cons, x), y)) → APP(cons, app(app(min, x), y))
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(app(min, x), app(app(cons, y), z)) → APP(if, app(app(le, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(del, x), app(app(cons, y), z)) → APP(app(del, x), z)
APP(app(del, x), app(app(cons, y), z)) → APP(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, y), z)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(del, x), app(app(cons, y), z)) → APP(app(eq, x), y)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(min, x), app(app(cons, y), z)) → APP(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
APP(minsort, app(app(cons, x), y)) → APP(app(del, app(app(min, x), y)), app(app(cons, x), y))
APP(app(del, x), app(app(cons, y), z)) → APP(app(if, app(app(eq, x), y)), z)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, x), z)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(del, x), app(app(cons, y), z)) → APP(app(cons, y), app(app(del, x), z))
APP(app(eq, app(s, x)), app(s, y)) → APP(eq, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(minsort, app(app(cons, x), y)) → APP(min, x)
APP(minsort, app(app(cons, x), y)) → APP(app(min, x), y)
APP(app(eq, app(s, x)), app(s, y)) → APP(app(eq, x), y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(minsort, app(app(cons, x), y)) → APP(del, app(app(min, x), y))
APP(app(min, x), app(app(cons, y), z)) → APP(le, x)
APP(app(del, x), app(app(cons, y), z)) → APP(if, app(app(eq, x), y))
APP(app(min, x), app(app(cons, y), z)) → APP(app(if, app(app(le, x), y)), app(app(min, x), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(minsort, app(app(cons, x), y)) → APP(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(min, x), app(app(cons, y), z)) → APP(min, y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(del, x), app(app(cons, y), z)) → APP(eq, x)
APP(minsort, app(app(cons, x), y)) → APP(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(min, x), app(app(cons, y), z)) → APP(app(le, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(minsort, app(app(cons, x), y)) → APP(cons, app(app(min, x), y))
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(app(min, x), app(app(cons, y), z)) → APP(if, app(app(le, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(del, x), app(app(cons, y), z)) → APP(app(del, x), z)
APP(app(del, x), app(app(cons, y), z)) → APP(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, y), z)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(del, x), app(app(cons, y), z)) → APP(app(eq, x), y)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(min, x), app(app(cons, y), z)) → APP(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
APP(minsort, app(app(cons, x), y)) → APP(app(del, app(app(min, x), y)), app(app(cons, x), y))
APP(app(del, x), app(app(cons, y), z)) → APP(app(if, app(app(eq, x), y)), z)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, x), z)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(del, x), app(app(cons, y), z)) → APP(app(cons, y), app(app(del, x), z))
APP(app(eq, app(s, x)), app(s, y)) → APP(eq, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(minsort, app(app(cons, x), y)) → APP(min, x)
APP(minsort, app(app(cons, x), y)) → APP(app(min, x), y)
APP(app(eq, app(s, x)), app(s, y)) → APP(app(eq, x), y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(minsort, app(app(cons, x), y)) → APP(del, app(app(min, x), y))
APP(app(min, x), app(app(cons, y), z)) → APP(le, x)
APP(app(del, x), app(app(cons, y), z)) → APP(if, app(app(eq, x), y))
APP(app(min, x), app(app(cons, y), z)) → APP(app(if, app(app(le, x), y)), app(app(min, x), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(minsort, app(app(cons, x), y)) → APP(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(min, x), app(app(cons, y), z)) → APP(min, y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

APP(app(del, x), app(app(cons, y), z)) → APP(eq, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(minsort, app(app(cons, x), y)) → APP(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(min, x), app(app(cons, y), z)) → APP(app(le, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(min, x), app(app(cons, y), z)) → APP(if, app(app(le, x), y))
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(minsort, app(app(cons, x), y)) → APP(cons, app(app(min, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(del, x), app(app(cons, y), z)) → APP(app(del, x), z)
APP(app(del, x), app(app(cons, y), z)) → APP(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, y), z)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(del, x), app(app(cons, y), z)) → APP(app(eq, x), y)
APP(minsort, app(app(cons, x), y)) → APP(app(del, app(app(min, x), y)), app(app(cons, x), y))
APP(app(min, x), app(app(cons, y), z)) → APP(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(del, x), app(app(cons, y), z)) → APP(app(if, app(app(eq, x), y)), z)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, x), z)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(del, x), app(app(cons, y), z)) → APP(app(cons, y), app(app(del, x), z))
APP(app(eq, app(s, x)), app(s, y)) → APP(eq, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(minsort, app(app(cons, x), y)) → APP(min, x)
APP(app(eq, app(s, x)), app(s, y)) → APP(app(eq, x), y)
APP(minsort, app(app(cons, x), y)) → APP(app(min, x), y)
APP(minsort, app(app(cons, x), y)) → APP(del, app(app(min, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(min, x), app(app(cons, y), z)) → APP(le, x)
APP(app(del, x), app(app(cons, y), z)) → APP(if, app(app(eq, x), y))
APP(app(min, x), app(app(cons, y), z)) → APP(app(if, app(app(le, x), y)), app(app(min, x), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(minsort, app(app(cons, x), y)) → APP(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(min, x), app(app(cons, y), z)) → APP(min, y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 29 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

APP(app(eq, app(s, x)), app(s, y)) → APP(app(eq, x), y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

APP(app(eq, app(s, x)), app(s, y)) → APP(app(eq, x), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2) = x2
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

APP(app(del, x), app(app(cons, y), z)) → APP(app(del, x), z)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

DEL(x, cons(y, z)) → DEL(x, z)

R is empty.
The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

APP(app(del, x), app(app(cons, y), z)) → APP(app(del, x), z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DEL(x1, x2) = x2
cons(x1, x2) = cons(x2)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

LE(s(x), s(y)) → LE(x, y)

R is empty.
The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LE(x1, x2) = x2
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

APP(app(min, x), app(app(cons, y), z)) → APP(app(min, x), z)
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, y), z)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

R is empty.
The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

APP(app(min, x), app(app(cons, y), z)) → APP(app(min, x), z)
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, y), z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MIN(x1, x2) = x2
cons(x1, x2) = cons(x2)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

APP(minsort, app(app(cons, x), y)) → APP(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y)))

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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.

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The remaining pairs can at least be oriented weakly.

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = x2
app(x1, x2) = app(x1, x2)
cons = cons

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.