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(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), 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(app(consif, true), x), ys) → APP(app(cons, x), ys)
APP(app(app(consif, true), x), ys) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(consif, app(f, x)), x), app(app(filter, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(consif, app(f, x)), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(consif, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), 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 5 less nodes.

(4) Obligation:

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

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(filter, f), app(app(cons, x), xs)) → APP(f, x)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(APP(x₁, x₂)) = x₁   
POL(app(x₁, x₂)) = 1 + x₁ + x₂   
POL(cons) = 0   
POL(filter) = 1   

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(APP(x₁, x₂)) = x₂   
POL(app(x₁, x₂)) = 1 + x₁ + x₂   
POL(cons) = 0   
POL(filter) = 0   

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

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

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) TRUE