Termination w.r.t. Q proof of /tmp/tmpQ8R1L5/012.trs

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:

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(forsome, p), app(app(cons, x), xs)) → APP(or, app(p, x))
APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forall, p), app(app(cons, x), xs)) → APP(app(and, app(p, x)), app(app(forall, p), xs))
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(or, app(p, x)), app(app(forsome, p), xs))
APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs)
APP(app(forall, p), app(app(cons, x), xs)) → APP(and, app(p, x))

The TRS R consists of the following rules:

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

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:

APP(app(forsome, p), app(app(cons, x), xs)) → APP(or, app(p, x))
APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forall, p), app(app(cons, x), xs)) → APP(app(and, app(p, x)), app(app(forall, p), xs))
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(or, app(p, x)), app(app(forsome, p), xs))
APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs)
APP(app(forall, p), app(app(cons, x), xs)) → APP(and, app(p, x))

The TRS R consists of the following rules:

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs)

The TRS R consists of the following rules:

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (2)x_1 + x_2
POL(cons) = 3
POL(forall) = 2
POL(app(x₁, x₂)) = (2)x_1 + x_2
POL(forsome) = 4

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

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.