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(i, 1) -> 1
app2(i, app2(i, x)) -> 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(i, 1) -> 1
app2(i, app2(i, x)) -> x

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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(i, 1) -> 1
app2(i, app2(i, x)) -> x

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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(i, 1) -> 1
app2(i, app2(i, x)) -> 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.