Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

app2(app2(neq, 0), 0) -> false
app2(app2(neq, 0), app2(s, y)) -> true
app2(app2(neq, app2(s, x)), 0) -> true
app2(app2(neq, app2(s, x)), app2(s, y)) -> app2(app2(neq, x), y)
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, y), ys)) -> app2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
app2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> app2(app2(cons, y), app2(app2(filter, f), ys))
app2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> app2(app2(filter, f), ys)
nonzero -> app2(filter, app2(neq, 0))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(app2(neq, 0), 0) -> false
app2(app2(neq, 0), app2(s, y)) -> true
app2(app2(neq, app2(s, x)), 0) -> true
app2(app2(neq, app2(s, x)), app2(s, y)) -> app2(app2(neq, x), y)
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, y), ys)) -> app2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
app2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> app2(app2(cons, y), app2(app2(filter, f), ys))
app2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> app2(app2(filter, f), ys)
nonzero -> app2(filter, app2(neq, 0))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

app2(app2(neq, 0), 0) -> false
app2(app2(neq, 0), app2(s, y)) -> true
app2(app2(neq, app2(s, x)), 0) -> true
app2(app2(neq, app2(s, x)), app2(s, y)) -> app2(app2(neq, x), y)
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, y), ys)) -> app2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
app2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> app2(app2(cons, y), app2(app2(filter, f), ys))
app2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> app2(app2(filter, f), ys)
nonzero -> app2(filter, app2(neq, 0))

The set Q consists of the following terms:

app2(app2(neq, 0), 0)
app2(app2(neq, 0), app2(s, x0))
app2(app2(neq, app2(s, x0)), 0)
app2(app2(neq, app2(s, x0)), app2(s, x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, true), x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, false), x0), app2(app2(cons, x1), x2))
nonzero


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

APP2(app2(neq, app2(s, x)), app2(s, y)) -> APP2(neq, x)
APP2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> APP2(app2(cons, y), app2(app2(filter, f), ys))
NONZERO -> APP2(neq, 0)
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(filtersub, app2(f, y))
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(app2(filtersub, app2(f, y)), f)
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(f, y)
APP2(app2(neq, app2(s, x)), app2(s, y)) -> APP2(app2(neq, x), y)
APP2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> APP2(filter, f)
APP2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> APP2(app2(filter, f), ys)
APP2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> APP2(filter, f)
APP2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> APP2(app2(filter, f), ys)
NONZERO -> APP2(filter, app2(neq, 0))

The TRS R consists of the following rules:

app2(app2(neq, 0), 0) -> false
app2(app2(neq, 0), app2(s, y)) -> true
app2(app2(neq, app2(s, x)), 0) -> true
app2(app2(neq, app2(s, x)), app2(s, y)) -> app2(app2(neq, x), y)
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, y), ys)) -> app2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
app2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> app2(app2(cons, y), app2(app2(filter, f), ys))
app2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> app2(app2(filter, f), ys)
nonzero -> app2(filter, app2(neq, 0))

The set Q consists of the following terms:

app2(app2(neq, 0), 0)
app2(app2(neq, 0), app2(s, x0))
app2(app2(neq, app2(s, x0)), 0)
app2(app2(neq, app2(s, x0)), app2(s, x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, true), x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, false), x0), app2(app2(cons, x1), x2))
nonzero

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

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

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

APP2(app2(neq, app2(s, x)), app2(s, y)) -> APP2(neq, x)
APP2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> APP2(app2(cons, y), app2(app2(filter, f), ys))
NONZERO -> APP2(neq, 0)
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(filtersub, app2(f, y))
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(app2(filtersub, app2(f, y)), f)
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(f, y)
APP2(app2(neq, app2(s, x)), app2(s, y)) -> APP2(app2(neq, x), y)
APP2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> APP2(filter, f)
APP2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> APP2(app2(filter, f), ys)
APP2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> APP2(filter, f)
APP2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> APP2(app2(filter, f), ys)
NONZERO -> APP2(filter, app2(neq, 0))

The TRS R consists of the following rules:

app2(app2(neq, 0), 0) -> false
app2(app2(neq, 0), app2(s, y)) -> true
app2(app2(neq, app2(s, x)), 0) -> true
app2(app2(neq, app2(s, x)), app2(s, y)) -> app2(app2(neq, x), y)
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, y), ys)) -> app2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
app2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> app2(app2(cons, y), app2(app2(filter, f), ys))
app2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> app2(app2(filter, f), ys)
nonzero -> app2(filter, app2(neq, 0))

The set Q consists of the following terms:

app2(app2(neq, 0), 0)
app2(app2(neq, 0), app2(s, x0))
app2(app2(neq, app2(s, x0)), 0)
app2(app2(neq, app2(s, x0)), app2(s, x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, true), x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, false), x0), app2(app2(cons, x1), x2))
nonzero

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 8 less nodes.

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

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

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

The TRS R consists of the following rules:

app2(app2(neq, 0), 0) -> false
app2(app2(neq, 0), app2(s, y)) -> true
app2(app2(neq, app2(s, x)), 0) -> true
app2(app2(neq, app2(s, x)), app2(s, y)) -> app2(app2(neq, x), y)
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, y), ys)) -> app2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
app2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> app2(app2(cons, y), app2(app2(filter, f), ys))
app2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> app2(app2(filter, f), ys)
nonzero -> app2(filter, app2(neq, 0))

The set Q consists of the following terms:

app2(app2(neq, 0), 0)
app2(app2(neq, 0), app2(s, x0))
app2(app2(neq, app2(s, x0)), 0)
app2(app2(neq, app2(s, x0)), app2(s, x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, true), x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, false), x0), app2(app2(cons, x1), x2))
nonzero

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(app2(neq, app2(s, x)), app2(s, y)) -> APP2(app2(neq, x), y)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app1(x2)
s  =  s
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

app2(app2(neq, 0), 0) -> false
app2(app2(neq, 0), app2(s, y)) -> true
app2(app2(neq, app2(s, x)), 0) -> true
app2(app2(neq, app2(s, x)), app2(s, y)) -> app2(app2(neq, x), y)
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, y), ys)) -> app2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
app2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> app2(app2(cons, y), app2(app2(filter, f), ys))
app2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> app2(app2(filter, f), ys)
nonzero -> app2(filter, app2(neq, 0))

The set Q consists of the following terms:

app2(app2(neq, 0), 0)
app2(app2(neq, 0), app2(s, x0))
app2(app2(neq, app2(s, x0)), 0)
app2(app2(neq, app2(s, x0)), app2(s, x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, true), x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, false), x0), app2(app2(cons, x1), x2))
nonzero

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

APP2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> APP2(app2(filter, f), ys)
APP2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> APP2(app2(filter, f), ys)
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(f, y)
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))

The TRS R consists of the following rules:

app2(app2(neq, 0), 0) -> false
app2(app2(neq, 0), app2(s, y)) -> true
app2(app2(neq, app2(s, x)), 0) -> true
app2(app2(neq, app2(s, x)), app2(s, y)) -> app2(app2(neq, x), y)
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, y), ys)) -> app2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
app2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> app2(app2(cons, y), app2(app2(filter, f), ys))
app2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> app2(app2(filter, f), ys)
nonzero -> app2(filter, app2(neq, 0))

The set Q consists of the following terms:

app2(app2(neq, 0), 0)
app2(app2(neq, 0), app2(s, x0))
app2(app2(neq, app2(s, x0)), 0)
app2(app2(neq, app2(s, x0)), app2(s, x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, true), x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, false), x0), app2(app2(cons, x1), x2))
nonzero

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> APP2(app2(filter, f), ys)
APP2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> APP2(app2(filter, f), ys)
APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(f, y)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app2(x1, x2)
cons  =  cons
Used ordering: Quasi Precedence: trivial 


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

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

APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))

The TRS R consists of the following rules:

app2(app2(neq, 0), 0) -> false
app2(app2(neq, 0), app2(s, y)) -> true
app2(app2(neq, app2(s, x)), 0) -> true
app2(app2(neq, app2(s, x)), app2(s, y)) -> app2(app2(neq, x), y)
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, y), ys)) -> app2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
app2(app2(app2(filtersub, true), f), app2(app2(cons, y), ys)) -> app2(app2(cons, y), app2(app2(filter, f), ys))
app2(app2(app2(filtersub, false), f), app2(app2(cons, y), ys)) -> app2(app2(filter, f), ys)
nonzero -> app2(filter, app2(neq, 0))

The set Q consists of the following terms:

app2(app2(neq, 0), 0)
app2(app2(neq, 0), app2(s, x0))
app2(app2(neq, app2(s, x0)), 0)
app2(app2(neq, app2(s, x0)), app2(s, x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, true), x0), app2(app2(cons, x1), x2))
app2(app2(app2(filtersub, false), x0), app2(app2(cons, x1), x2))
nonzero

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.