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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 DependencyGraphProof (⇔)
        ↳14 TRUE

(0) Obligation:

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

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), 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(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(app(copy, n), y), z)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(copy, n), y)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(copy, n)
APP(app(app(copy, 0), y), z) → APP(f, z)
APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(app(copy, app(s, x)), y), z) → APP(app(copy, x), y)
APP(app(app(copy, app(s, x)), y), z) → APP(copy, x)
APP(app(app(copy, app(s, x)), y), z) → APP(app(cons, app(f, y)), z)
APP(app(app(copy, app(s, x)), y), z) → APP(cons, app(f, y))
APP(app(app(copy, app(s, x)), y), z) → APP(f, y)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)

The TRS R consists of the following rules:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), 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 2 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))

The TRS R consists of the following rules:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
APP(x0, x1, x2)  =  APP(x1)

Tags:
APP has argument tags [0,0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x2)
app(x1, x2)  =  app(x1, x2)
copy  =  copy
s  =  s
cons  =  cons
f  =  f
nil  =  nil
n  =  n

Recursive path order with status [RPO].
Quasi-Precedence:
[f, n] > app2 > copy > [APP1, s] > cons
nil > copy > [APP1, s] > cons

Status:
APP1: multiset
app2: [2,1]
copy: multiset
s: multiset
cons: multiset
f: multiset
nil: multiset
n: multiset


The following usable rules [FROCOS05] were oriented:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)

(7) Obligation:

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

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)

The TRS R consists of the following rules:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
APP(x0, x1, x2)  =  APP(x0, x2)

Tags:
APP has argument tags [1,1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
APP(x1, x2)  =  APP
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
filter  =  filter
filter2  =  filter2
true  =  true
false  =  false
f  =  f
nil  =  nil
copy  =  copy
n  =  n
0  =  0
s  =  s

Recursive path order with status [RPO].
Quasi-Precedence:
filter2 > [APP, app2, cons, f, copy] > filter > nil > map
true > map
false > map
n > map
0 > map
s > map

Status:
APP: multiset
app2: multiset
map: multiset
cons: multiset
filter: multiset
filter2: multiset
true: multiset
false: multiset
f: multiset
nil: multiset
copy: multiset
n: multiset
0: multiset
s: multiset


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)

The TRS R consists of the following rules:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

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

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(14) TRUE