Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(4) Obligation:

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

(5) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
forall  =  forall
cons  =  cons
forsome  =  forsome
app(x1, x2)  =  app(x2)

From the DPs we obtained the following set of size-change graphs:

• APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1

• APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1

• APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(6) TRUE