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


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(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 [13,14,18] contains 2 SCCs with 3 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(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), 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(x1)
app2(x1, x2)  =  app2(x1, x2)
plus  =  plus
s  =  s

Lexicographic Path Order [19].
Precedence:
APP1 > app2
plus > app2
s > 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(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
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(app2(curry, f), x), y) -> APP2(app2(f, x), y)
APP2(app2(app2(curry, f), x), y) -> 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)  =  x1
app2(x1, x2)  =  app2(x1, x2)
curry  =  curry
plus  =  plus
0  =  0
s  =  s

Lexicographic Path Order [19].
Precedence:
curry > s
plus > app2 > s
0 > s

The following usable rules [14] were oriented:

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)



↳ 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(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.