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 DependencyGraphProof (⇔)
    ↳16 TRUE

(0) Obligation:

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

app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
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) 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(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
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)

The set Q consists of the following terms:

app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
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(perfectp, app(s, x)) → APP(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
APP(perfectp, app(s, x)) → APP(app(app(f, x), app(s, 0)), app(s, x))
APP(perfectp, app(s, x)) → APP(app(f, x), app(s, 0))
APP(perfectp, app(s, x)) → APP(f, x)
APP(perfectp, app(s, x)) → APP(s, 0)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(f, x), u), app(app(minus, z), app(s, x)))
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(f, x), u)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(f, x)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(minus, z), app(s, x))
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(minus, z)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(if, app(app(le, x), y))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(le, x), y)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(le, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(f, app(s, x)), app(app(minus, y), x)), z)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(f, app(s, x)), app(app(minus, y), x))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(minus, y), x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(minus, y)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, x), u), z), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(f, x), u), z)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(f, x), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(f, x)
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(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
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)

The set Q consists of the following terms:

app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
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 2 SCCs with 32 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, x), u), z), u)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)

The TRS R consists of the following rules:

app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
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)

The set Q consists of the following terms:

app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
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

f1(s(x), s(y), z, u) → f1(x, u, z, u)
f1(s(x), 0, z, u) → f1(x, u, minus(z, s(x)), u)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, x), u), z), u)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
f1(x1, x2, x3, x4)  =  f1(x1, x3, x4)
s(x1)  =  s(x1)
0  =  0
minus(x1, x2)  =  minus

Lexicographic Path Order [LPO].
Precedence:
f13 > minus > s1
0 > s1

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(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
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)

The set Q consists of the following terms:

app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
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(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(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
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)

The set Q consists of the following terms:

app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
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].


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: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
filter  =  filter
filter2  =  filter2
true  =  true
false  =  false
f  =  f
0  =  0
perfectp  =  perfectp
s  =  s
minus  =  minus
nil  =  nil
if  =  if
le  =  le

Lexicographic Path Order [LPO].
Precedence:
map > app2
cons > app2
filter > app2
filter2 > app2
true > app2
false > app2

The following usable rules [FROCOS05] were oriented: none

(14) 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(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
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)

The set Q consists of the following terms:

app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
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) DependencyGraphProof (EQUIVALENT transformation)

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

(16) TRUE