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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 PisEmptyProof (⇔)
    ↳23 TRUE

(0) Obligation:

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

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

APP(rev, app(app(cons, x), l)) → APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
APP(rev, app(app(cons, x), l)) → APP(cons, app(app(rev1, x), l))
APP(rev, app(app(cons, x), l)) → APP(app(rev1, x), l)
APP(rev, app(app(cons, x), l)) → APP(rev1, x)
APP(rev, app(app(cons, x), l)) → APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) → APP(rev2, x)
APP(app(rev1, x), app(app(cons, y), l)) → APP(app(rev1, y), l)
APP(app(rev1, x), app(app(cons, y), l)) → APP(rev1, y)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev2, x), app(app(cons, y), l)) → APP(app(cons, x), app(app(rev2, y), l))
APP(app(rev2, x), app(app(cons, y), l)) → APP(cons, x)
APP(app(rev2, x), app(app(cons, y), l)) → APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev2, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)

The TRS R consists of the following rules:

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 19 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(rev1, x), app(app(cons, y), l)) → APP(app(rev1, y), l)

The TRS R consists of the following rules:

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

rev11(x, cons(y, l)) → rev11(y, l)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(rev1, x), app(app(cons, y), l)) → APP(app(rev1, y), l)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
cons2 > rev112

Status:
cons2: [1,2]
rev112: [1,2]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

APP(rev, app(app(cons, x), l)) → APP(app(rev2, x), l)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev2, x), app(app(cons, y), l)) → APP(app(rev2, y), l)

The TRS R consists of the following rules:

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

rev1(cons(x, l)) → rev21(x, l)
rev21(x, cons(y, l)) → rev1(cons(x, rev2(y, l)))
rev21(x, cons(y, l)) → rev21(y, l)

The a-transformed usable rules are

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)


The following pairs can be oriented strictly and are deleted.


APP(rev, app(app(cons, x), l)) → APP(app(rev2, x), l)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev2, x), app(app(cons, y), l)) → APP(app(rev2, y), l)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
rev1(x1)  =  rev1(x1)
cons(x1, x2)  =  cons(x2)
rev21(x1, x2)  =  rev21(x2)
rev2(x1, x2)  =  x2
nil  =  nil
rev(x1)  =  x1
rev1(x1, x2)  =  x2
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Precedence:
cons1 > rev211 > rev11
nil > 0 > rev11
nil > s > rev11

Status:
rev211: [1]
cons1: [1]
rev11: [1]
s: []
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)

(14) Obligation:

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

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

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

The TRS R consists of the following rules:

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
filter  =  filter
filter2  =  filter2
true  =  true
false  =  false

Lexicographic path order with status [LPO].
Precedence:
map > app2 > filter
cons > filter
filter2 > filter
true > app2 > filter
false > app2 > filter

Status:
cons: []
true: []
map: []
false: []
app2: [2,1]
filter2: []
filter: []

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

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

The TRS R consists of the following rules:

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

map1(f, cons(x, xs)) → map1(f, xs)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
cons2 > map12

Status:
cons2: [1,2]
map12: [1,2]

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE