Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) → app(app(g, x), y)
inc → app(map, app(app(curry, plus), app(s, 0)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) → app(app(g, x), y)
inc → app(map, app(app(curry, plus), app(s, 0)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
INC → APP(map, app(app(curry, plus), app(s, 0)))
INC → APP(curry, plus)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
INC → APP(app(curry, plus), app(s, 0))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(app(curry, g), x), y) → APP(g, x)
INC → APP(s, 0)
APP(app(app(curry, g), x), y) → APP(app(g, x), y)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) → app(app(g, x), y)
inc → app(map, app(app(curry, plus), app(s, 0)))

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:

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
INC → APP(map, app(app(curry, plus), app(s, 0)))
INC → APP(curry, plus)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
INC → APP(app(curry, plus), app(s, 0))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(app(curry, g), x), y) → APP(g, x)
INC → APP(s, 0)
APP(app(app(curry, g), x), y) → APP(app(g, x), y)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) → app(app(g, x), y)
inc → app(map, app(app(curry, plus), app(s, 0)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 8 less nodes.

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

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

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) → app(app(g, x), y)
inc → app(map, app(app(curry, plus), app(s, 0)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x1, x2)) = (1/4)x_1   
POL(plus) = 0   
POL(app(x1, x2)) = 1 + x_2   
POL(s) = 0   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] 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:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) → app(app(g, x), y)
inc → app(map, app(app(curry, plus), app(s, 0)))

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

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

APP(app(app(curry, g), x), y) → APP(g, x)
APP(app(app(curry, g), x), y) → APP(app(g, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) → app(app(g, x), y)
inc → app(map, app(app(curry, plus), app(s, 0)))

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