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.

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(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 [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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 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)
.  =  .
1  =  1
i  =  i

Lexicographic Path Order [19].
Precedence:
[APP1, ., 1] > app2
i > app2


The following usable rules [14] were oriented:

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))



↳ 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(., 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
            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(i, app2(app2(., x), y)) -> APP2(i, y)
APP2(i, app2(app2(., x), y)) -> APP2(i, 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)  =  APP1(x2)
i  =  i
app2(x1, x2)  =  app2(x1, x2)
.  =  .

Lexicographic Path Order [19].
Precedence:
app2 > APP1


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(., 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.