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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)

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(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(filtersub, app(f, y)), f)
APP(app(filter, f), app(app(cons, y), ys)) → APP(filtersub, app(f, y))
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(cons, y), app(app(filter, f), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(filter, f)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(filter, f)

The TRS R consists of the following rules:

app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)

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 6 less nodes.

(4) Obligation:

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

APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x2)
app(x1, x2)  =  app(x1, x2)
filtersub  =  filtersub
true  =  true
cons  =  cons
filter  =  filter
false  =  false

Lexicographic path order with status [LPO].
Quasi-Precedence:

filter > [APP₁, true, cons] > [app₂, filtersub, false]

Status:

APP₁: [1]
app₂: [2,1]
filtersub: []
true: []
cons: []
filter: []
false: []

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE