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
  ↳ 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))

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:

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))

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

↳ 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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))

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.


APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(reverse2, xs), app2(app2(cons, x), l))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1   
POL(app2(x1, x2)) = 1 + x1 + x2   
POL(cons) = 0   
POL(reverse2) = 0   

The following usable rules [14] were oriented: none



↳ 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(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.
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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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))

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.


APP2(app2(app2(compose, f), g), x) -> APP2(g, app2(f, x))
APP2(app2(app2(compose, f), g), x) -> APP2(f, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1   
POL(app2(x1, x2)) = 1 + x1 + x2   
POL(compose) = 1   
POL(cons) = 0   
POL(hd) = 0   
POL(nil) = 0   
POL(reverse) = 0   
POL(reverse2) = 0   
POL(tl) = 0   

The following usable rules [14] were oriented: none



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