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(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))

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(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))

The set Q consists of the following terms:

app2(f, app2(s, x0))
app2(g, app2(app2(cons, 0), x0))
app2(g, app2(app2(cons, app2(s, x0)), x1))
app2(h, app2(app2(cons, x0), x1))


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(g, app2(app2(cons, 0), y)) -> APP2(g, y)
APP2(f, app2(s, x)) -> APP2(f, x)
APP2(h, app2(app2(cons, x), y)) -> APP2(g, app2(app2(cons, x), y))
APP2(h, app2(app2(cons, x), y)) -> APP2(h, app2(g, app2(app2(cons, x), y)))

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))

The set Q consists of the following terms:

app2(f, app2(s, x0))
app2(g, app2(app2(cons, 0), x0))
app2(g, app2(app2(cons, app2(s, x0)), x1))
app2(h, app2(app2(cons, x0), x1))

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(g, app2(app2(cons, 0), y)) -> APP2(g, y)
APP2(f, app2(s, x)) -> APP2(f, x)
APP2(h, app2(app2(cons, x), y)) -> APP2(g, app2(app2(cons, x), y))
APP2(h, app2(app2(cons, x), y)) -> APP2(h, app2(g, app2(app2(cons, x), y)))

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))

The set Q consists of the following terms:

app2(f, app2(s, x0))
app2(g, app2(app2(cons, 0), x0))
app2(g, app2(app2(cons, app2(s, x0)), x1))
app2(h, app2(app2(cons, x0), x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 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(g, app2(app2(cons, 0), y)) -> APP2(g, y)

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))

The set Q consists of the following terms:

app2(f, app2(s, x0))
app2(g, app2(app2(cons, 0), x0))
app2(g, app2(app2(cons, app2(s, x0)), x1))
app2(h, app2(app2(cons, x0), x1))

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(g, app2(app2(cons, 0), y)) -> APP2(g, 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)  =  APP1(x2)
g  =  g
app2(x1, x2)  =  app2(x1, x2)
cons  =  cons
0  =  0

Lexicographic Path Order [19].
Precedence:
[APP1, app2]


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(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))

The set Q consists of the following terms:

app2(f, app2(s, x0))
app2(g, app2(app2(cons, 0), x0))
app2(g, app2(app2(cons, app2(s, x0)), x1))
app2(h, app2(app2(cons, x0), x1))

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(f, app2(s, x)) -> APP2(f, x)

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))

The set Q consists of the following terms:

app2(f, app2(s, x0))
app2(g, app2(app2(cons, 0), x0))
app2(g, app2(app2(cons, app2(s, x0)), x1))
app2(h, app2(app2(cons, x0), x1))

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(f, app2(s, x)) -> APP2(f, x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
f  =  f
app2(x1, x2)  =  app1(x2)
s  =  s

Lexicographic Path Order [19].
Precedence:
[APP1, app1] > f
s > f


The following usable rules [14] were oriented: none



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

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

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))

The set Q consists of the following terms:

app2(f, app2(s, x0))
app2(g, app2(app2(cons, 0), x0))
app2(g, app2(app2(cons, app2(s, x0)), x1))
app2(h, app2(app2(cons, x0), x1))

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