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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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(fact, 0) → APP(s, 0)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), 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, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(*, app(s, x)), y) → APP(app(*, x), y)
APP(app(*, app(s, x)), y) → APP(app(+, app(app(*, x), y)), y)
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(fact, app(s, x)) → APP(p, app(s, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(*, app(s, x)), y) → APP(+, app(app(*, x), y))
APP(app(+, x), app(s, y)) → APP(app(+, x), y)
APP(fact, app(s, x)) → APP(*, app(s, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(fact, app(s, x)) → APP(fact, app(p, app(s, 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(app(*, app(s, x)), y) → APP(*, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(fact, app(s, x)) → APP(app(*, app(s, x)), app(fact, app(p, app(s, x))))
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(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(+, x), app(s, y)) → APP(s, app(app(+, x), y))

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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(fact, 0) → APP(s, 0)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), 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, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(*, app(s, x)), y) → APP(app(*, x), y)
APP(app(*, app(s, x)), y) → APP(app(+, app(app(*, x), y)), y)
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(fact, app(s, x)) → APP(p, app(s, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(*, app(s, x)), y) → APP(+, app(app(*, x), y))
APP(app(+, x), app(s, y)) → APP(app(+, x), y)
APP(fact, app(s, x)) → APP(*, app(s, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(fact, app(s, x)) → APP(fact, app(p, app(s, 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(app(*, app(s, x)), y) → APP(*, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(fact, app(s, x)) → APP(app(*, app(s, x)), app(fact, app(p, app(s, x))))
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(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(+, x), app(s, y)) → APP(s, app(app(+, x), y))

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(fact, 0) → APP(s, 0)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), 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, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(*, app(s, x)), y) → APP(app(*, x), y)
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(*, app(s, x)), y) → APP(app(+, app(app(*, x), y)), y)
APP(fact, app(s, x)) → APP(p, app(s, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(*, app(s, x)), y) → APP(+, app(app(*, x), y))
APP(app(+, x), app(s, y)) → APP(app(+, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(fact, app(s, x)) → APP(*, app(s, x))
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(fact, app(s, x)) → APP(fact, app(p, app(s, 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(app(*, app(s, x)), y) → APP(*, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(fact, app(s, x)) → APP(app(*, app(s, x)), app(fact, app(p, app(s, x))))
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(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(+, x), app(s, y)) → APP(s, app(app(+, x), y))

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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 4 SCCs with 17 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

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

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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:

+1(x, s(y)) → +1(x, y)

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

P(s(x0))
FACT(0)
FACT(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(x1))
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(+, x), app(s, y)) → APP(app(+, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+1(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

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

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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

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

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

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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:

*1(s(x), y) → *1(x, y)

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

P(s(x0))
FACT(0)
FACT(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(x1))
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(*, app(s, x)), y) → APP(app(*, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  x1
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
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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

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

APP(fact, app(s, x)) → APP(fact, app(p, app(s, x)))

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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
                    ↳ 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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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
                    ↳ 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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
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.