Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

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(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(app2(until, x0), x1), x2)


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(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
APP2(app2(app2(until, p), f), x) -> APP2(app2(if, app2(p, x)), x)
APP2(app2(app2(until, p), f), x) -> APP2(if, app2(p, x))
APP2(app2(app2(until, p), f), x) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(app2(until, x0), x1), x2)

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(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
APP2(app2(app2(until, p), f), x) -> APP2(app2(if, app2(p, x)), x)
APP2(app2(app2(until, p), f), x) -> APP2(if, app2(p, x))
APP2(app2(app2(until, p), f), x) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(app2(until, x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

APP2(app2(app2(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
APP2(app2(app2(until, p), f), x) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(app2(until, x0), x1), x2)

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(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(f, x)
The remaining pairs can at least by weakly be oriented.

APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
until  =  until
if  =  if
true  =  true
false  =  false

Lexicographic Path Order [19].
Precedence:
until > app2
if > app2
true > app2
false > app2


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP

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

APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(app2(until, x0), x1), x2)

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