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


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP2(app2(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 [13,14,18] contains 2 SCCs with 8 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(f, y)
The remaining pairs can at least by weakly be 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(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x1)
app2(x1, x2)  =  app1(x2)
filtersub  =  filtersub
false  =  false
cons  =  cons
filter  =  filter
true  =  true
neq  =  neq
0  =  0
s  =  s
nil  =  nil

Lexicographic Path Order [19].
Precedence:
[filter, nil] > [APP1, filtersub] > [app1, cons, true] > false
neq > [app1, cons, true] > false
0 > [app1, cons, true] > false


The following usable rules [14] were oriented:

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



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

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(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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)
The remaining pairs can at least by weakly be oriented.

APP2(app2(filter, f), app2(app2(cons, y), ys)) -> APP2(app2(app2(filtersub, app2(f, y)), f), app2(app2(cons, y), ys))
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x2
app2(x1, x2)  =  app1(x2)
filtersub  =  filtersub
false  =  false
cons  =  cons
filter  =  filter
true  =  true
neq  =  neq
0  =  0
s  =  s
nil  =  nil

Lexicographic Path Order [19].
Precedence:
filtersub > cons > [app1, true, 0] > [false, neq] > filter
s > filter
nil > filter


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ 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 [13,14,18] contains 0 SCCs with 1 less node.