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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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)
The remaining pairs can at least be oriented weakly.

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)
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x1, x2)) = (2)x_1   
POL(cons) = 11/4   
POL(forall) = 4   
POL(app(x1, x2)) = 3/2 + (2)x_2   
POL(forsome) = 7/2   
The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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(app(forsome, p), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x1, x2)) = (7/2)x_2   
POL(cons) = 1/4   
POL(app(x1, x2)) = (1/4)x_1 + (7/2)x_2   
POL(forsome) = 0   
The value of delta used in the strict ordering is 7/128.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

app(app(and, true), true) → true
app(app(and, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

app(app(and, true), true) → true
app(app(and, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x1, x2)) = (7/2)x_2   
POL(cons) = 1/4   
POL(forall) = 0   
POL(app(x1, x2)) = (1/2)x_1 + (7/2)x_2   
The value of delta used in the strict ordering is 7/32.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ 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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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.