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(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

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(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init


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

LAST -> APP2(app2(compose, hd), reverse)
LAST -> APP2(compose, hd)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(cons, x), l)
INIT -> APP2(app2(compose, tl), reverse)
INIT -> APP2(compose, tl)
APP2(app2(app2(compose, f), g), x) -> APP2(f, x)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(reverse2, xs)
INIT -> APP2(compose, reverse)
APP2(reverse, l) -> APP2(reverse2, l)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(reverse2, xs), app2(app2(cons, x), l))
APP2(app2(app2(compose, f), g), x) -> APP2(g, app2(f, x))
APP2(reverse, l) -> APP2(app2(reverse2, l), nil)
INIT -> APP2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The TRS R consists of the following rules:

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

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:

LAST -> APP2(app2(compose, hd), reverse)
LAST -> APP2(compose, hd)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(cons, x), l)
INIT -> APP2(app2(compose, tl), reverse)
INIT -> APP2(compose, tl)
APP2(app2(app2(compose, f), g), x) -> APP2(f, x)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(reverse2, xs)
INIT -> APP2(compose, reverse)
APP2(reverse, l) -> APP2(reverse2, l)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(reverse2, xs), app2(app2(cons, x), l))
APP2(app2(app2(compose, f), g), x) -> APP2(g, app2(f, x))
APP2(reverse, l) -> APP2(app2(reverse2, l), nil)
INIT -> APP2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The TRS R consists of the following rules:

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 10 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(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(reverse2, xs), app2(app2(cons, x), l))

The TRS R consists of the following rules:

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

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(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(reverse2, xs), app2(app2(cons, x), l))
Used argument filtering: APP2(x1, x2)  =  x1
app2(x1, x2)  =  app1(x2)
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(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

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(compose, f), g), x) -> APP2(g, app2(f, x))
APP2(app2(app2(compose, f), g), x) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

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(compose, f), g), x) -> APP2(g, app2(f, x))
APP2(app2(app2(compose, f), g), x) -> APP2(f, x)
Used argument filtering: APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
compose  =  compose
Used ordering: Quasi Precedence: trivial 


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

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

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

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