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(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)

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(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)

The set Q consists of the following terms:

app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add


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

APP2(app2(app2(curry, f), x), y) -> APP2(app2(f, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(app2(curry, f), x), y) -> APP2(f, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
ADD -> APP2(curry, plus)

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)

The set Q consists of the following terms:

app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add

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(app2(curry, f), x), y) -> APP2(app2(f, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(app2(curry, f), x), y) -> APP2(f, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
ADD -> APP2(curry, plus)

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)

The set Q consists of the following terms:

app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 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(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)

The set Q consists of the following terms:

app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add

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(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
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(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)

The set Q consists of the following terms:

app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add

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(curry, f), x), y) -> APP2(app2(f, x), y)
APP2(app2(app2(curry, f), x), y) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)

The set Q consists of the following terms:

app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add

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(curry, f), x), y) -> APP2(app2(f, x), y)
APP2(app2(app2(curry, f), x), y) -> APP2(f, x)
Used argument filtering: APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
curry  =  curry
plus  =  plus
0  =  0
s  =  s
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(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)

The set Q consists of the following terms:

app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add

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