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(., 1), x) -> x
app2(app2(., x), 1) -> x
app2(app2(., app2(i, x)), x) -> 1
app2(app2(., x), app2(i, x)) -> 1
app2(app2(., app2(i, y)), app2(app2(., y), z)) -> z
app2(app2(., y), app2(app2(., app2(i, y)), z)) -> z
app2(app2(., app2(app2(., x), y)), z) -> app2(app2(., x), app2(app2(., y), z))
app2(i, 1) -> 1
app2(i, app2(i, x)) -> x
app2(i, app2(app2(., x), y)) -> app2(app2(., app2(i, y)), app2(i, x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(., 1), x) -> x
app2(app2(., x), 1) -> x
app2(app2(., app2(i, x)), x) -> 1
app2(app2(., x), app2(i, x)) -> 1
app2(app2(., app2(i, y)), app2(app2(., y), z)) -> z
app2(app2(., y), app2(app2(., app2(i, y)), z)) -> z
app2(app2(., app2(app2(., x), y)), z) -> app2(app2(., x), app2(app2(., y), z))
app2(i, 1) -> 1
app2(i, app2(i, x)) -> x
app2(i, app2(app2(., x), y)) -> app2(app2(., app2(i, y)), app2(i, x))

Q is empty.

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

APP2(i, app2(app2(., x), y)) -> APP2(i, y)
APP2(app2(., app2(app2(., x), y)), z) -> APP2(app2(., x), app2(app2(., y), z))
APP2(i, app2(app2(., x), y)) -> APP2(i, x)
APP2(i, app2(app2(., x), y)) -> APP2(., app2(i, y))
APP2(i, app2(app2(., x), y)) -> APP2(app2(., app2(i, y)), app2(i, x))
APP2(app2(., app2(app2(., x), y)), z) -> APP2(., y)
APP2(app2(., app2(app2(., x), y)), z) -> APP2(app2(., y), z)

The TRS R consists of the following rules:

app2(app2(., 1), x) -> x
app2(app2(., x), 1) -> x
app2(app2(., app2(i, x)), x) -> 1
app2(app2(., x), app2(i, x)) -> 1
app2(app2(., app2(i, y)), app2(app2(., y), z)) -> z
app2(app2(., y), app2(app2(., app2(i, y)), z)) -> z
app2(app2(., app2(app2(., x), y)), z) -> app2(app2(., x), app2(app2(., y), z))
app2(i, 1) -> 1
app2(i, app2(i, x)) -> x
app2(i, app2(app2(., x), y)) -> app2(app2(., app2(i, y)), app2(i, x))

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:

APP2(i, app2(app2(., x), y)) -> APP2(i, y)
APP2(app2(., app2(app2(., x), y)), z) -> APP2(app2(., x), app2(app2(., y), z))
APP2(i, app2(app2(., x), y)) -> APP2(i, x)
APP2(i, app2(app2(., x), y)) -> APP2(., app2(i, y))
APP2(i, app2(app2(., x), y)) -> APP2(app2(., app2(i, y)), app2(i, x))
APP2(app2(., app2(app2(., x), y)), z) -> APP2(., y)
APP2(app2(., app2(app2(., x), y)), z) -> APP2(app2(., y), z)

The TRS R consists of the following rules:

app2(app2(., 1), x) -> x
app2(app2(., x), 1) -> x
app2(app2(., app2(i, x)), x) -> 1
app2(app2(., x), app2(i, x)) -> 1
app2(app2(., app2(i, y)), app2(app2(., y), z)) -> z
app2(app2(., y), app2(app2(., app2(i, y)), z)) -> z
app2(app2(., app2(app2(., x), y)), z) -> app2(app2(., x), app2(app2(., y), z))
app2(i, 1) -> 1
app2(i, app2(i, x)) -> x
app2(i, app2(app2(., x), y)) -> app2(app2(., app2(i, y)), app2(i, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

APP2(app2(., app2(app2(., x), y)), z) -> APP2(app2(., x), app2(app2(., y), z))
APP2(app2(., app2(app2(., x), y)), z) -> APP2(app2(., y), z)

The TRS R consists of the following rules:

app2(app2(., 1), x) -> x
app2(app2(., x), 1) -> x
app2(app2(., app2(i, x)), x) -> 1
app2(app2(., x), app2(i, x)) -> 1
app2(app2(., app2(i, y)), app2(app2(., y), z)) -> z
app2(app2(., y), app2(app2(., app2(i, y)), z)) -> z
app2(app2(., app2(app2(., x), y)), z) -> app2(app2(., x), app2(app2(., y), z))
app2(i, 1) -> 1
app2(i, app2(i, x)) -> x
app2(i, app2(app2(., x), y)) -> app2(app2(., app2(i, y)), app2(i, x))

Q is empty.
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(app2(., x), y)), z) -> APP2(app2(., x), app2(app2(., y), z))
APP2(app2(., app2(app2(., x), y)), z) -> APP2(app2(., y), z)
Used argument filtering: APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
.  =  .
1  =  1
i  =  i
Used ordering: Quasi Precedence: app_2 > 1 . > 1 i > 1 


↳ 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(., 1), x) -> x
app2(app2(., x), 1) -> x
app2(app2(., app2(i, x)), x) -> 1
app2(app2(., x), app2(i, x)) -> 1
app2(app2(., app2(i, y)), app2(app2(., y), z)) -> z
app2(app2(., y), app2(app2(., app2(i, y)), z)) -> z
app2(app2(., app2(app2(., x), y)), z) -> app2(app2(., x), app2(app2(., y), z))
app2(i, 1) -> 1
app2(i, app2(i, x)) -> x
app2(i, app2(app2(., x), y)) -> app2(app2(., app2(i, y)), app2(i, x))

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
            ↳ QDPAfsSolverProof

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

APP2(i, app2(app2(., x), y)) -> APP2(i, y)
APP2(i, app2(app2(., x), y)) -> APP2(i, x)

The TRS R consists of the following rules:

app2(app2(., 1), x) -> x
app2(app2(., x), 1) -> x
app2(app2(., app2(i, x)), x) -> 1
app2(app2(., x), app2(i, x)) -> 1
app2(app2(., app2(i, y)), app2(app2(., y), z)) -> z
app2(app2(., y), app2(app2(., app2(i, y)), z)) -> z
app2(app2(., app2(app2(., x), y)), z) -> app2(app2(., x), app2(app2(., y), z))
app2(i, 1) -> 1
app2(i, app2(i, x)) -> x
app2(i, app2(app2(., x), y)) -> app2(app2(., app2(i, y)), app2(i, x))

Q is empty.
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(i, app2(app2(., x), y)) -> APP2(i, y)
APP2(i, app2(app2(., x), y)) -> APP2(i, x)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app2(x1, x2)
.  =  .
Used ordering: Quasi Precedence: trivial 


↳ 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(., 1), x) -> x
app2(app2(., x), 1) -> x
app2(app2(., app2(i, x)), x) -> 1
app2(app2(., x), app2(i, x)) -> 1
app2(app2(., app2(i, y)), app2(app2(., y), z)) -> z
app2(app2(., y), app2(app2(., app2(i, y)), z)) -> z
app2(app2(., app2(app2(., x), y)), z) -> app2(app2(., x), app2(app2(., y), z))
app2(i, 1) -> 1
app2(i, app2(i, x)) -> x
app2(i, app2(app2(., x), y)) -> app2(app2(., app2(i, y)), app2(i, x))

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.