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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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
  ↳ DependencyPairsProof

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.

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(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
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(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), 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(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
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(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(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(plus, app(app(minus, y), app(s, app(s, z))))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(plus, app(app(plus, y), app(s, app(s, z))))
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))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
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(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), 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(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
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(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(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(plus, app(app(minus, y), app(s, app(s, z))))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(plus, app(app(plus, y), app(s, app(s, z))))
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))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
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(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), y))
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
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(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(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(plus, app(app(minus, y), app(s, app(s, z))))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(plus, app(app(plus, y), app(s, app(s, z))))
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))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 18 less nodes.

↳ 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(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
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:

PLUS(minus(x, s(0)), minus(y, s(s(z)))) → PLUS(minus(y, s(s(z))), minus(x, s(0)))
PLUS(s(x), y) → PLUS(x, y)
PLUS(plus(x, s(0)), plus(y, s(s(z)))) → PLUS(plus(y, s(s(z))), plus(x, s(0)))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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


The following pairs can be oriented strictly and are deleted.


APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
The remaining pairs can at least be oriented weakly.

APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x1, x2)
minus(x1, x2)  =  minus(x1)
s(x1)  =  s(x1)
0  =  0
plus(x1, x2)  =  plus(x1, x2)

Recursive path order with status [2].
Precedence:
0 > plus2 > PLUS2 > s1
0 > plus2 > minus1 > s1

Status:
minus1: [1]
plus2: multiset
s1: multiset
0: multiset
PLUS2: multiset

The following usable rules [14] were oriented:

app(app(plus, 0), y) → y
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
We have to consider all minimal (P,Q,R)-chains.

↳ 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(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
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:

MINUS(s(x), s(y)) → MINUS(x, y)

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


The following pairs can be oriented strictly and are deleted.


APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x1)
s(x1)  =  s(x1)

Recursive path order with status [2].
Precedence:
s1 > MINUS1

Status:
MINUS1: multiset
s1: multiset

The following usable rules [14] were oriented: none



↳ 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(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP

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

APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

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


The following pairs can be oriented strictly and are deleted.


APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
0  =  0

Recursive path order with status [2].
Precedence:
s1 > QUOT1

Status:
QUOT1: multiset
s1: multiset
0: multiset

The following usable rules [14] were oriented:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ 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(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
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)  =  APP(x2)
app(x1, x2)  =  app(x1, x2)
filter2  =  filter2
true  =  true
filter  =  filter
map  =  map
cons  =  cons
false  =  false
nil  =  nil
minus  =  minus
s  =  s
quot  =  quot
plus  =  plus
0  =  0

Recursive path order with status [2].
Precedence:
filter2 > filter > app2 > APP1 > map
filter2 > filter > app2 > minus > plus
filter2 > filter > app2 > minus > 0
filter2 > cons > app2 > APP1 > map
filter2 > cons > app2 > minus > plus
filter2 > cons > app2 > minus > 0
quot > s > app2 > APP1 > map
quot > s > app2 > minus > plus
quot > s > app2 > minus > 0

Status:
APP1: multiset
map: multiset
0: multiset
s: multiset
minus: multiset
nil: multiset
cons: multiset
filter: multiset
true: multiset
false: multiset
app2: [2,1]
plus: multiset
filter2: multiset
quot: multiset

The following usable rules [14] were oriented: none



↳ 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(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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.
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.