Termination w.r.t. Q proof of /tmp/tmpB4jiWb/perfect2.trs

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 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 DependencyGraphProof (⇔)

    ↳26 TRUE

(0) Obligation:

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

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
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(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
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(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
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 4 SCCs with 33 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
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

le1(s(x), s(y)) → le1(x, y)

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
le1(x1, x2) = le1(x2)
s(x1) = s(x1)

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

s₁ > le1₁

Status:

le1₁: [1]
s₁: [1]

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(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
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(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
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)

minus1(s(x), s(y)) → minus1(x, y)

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
minus1(x1, x2) = minus1(x2)
s(x1) = s(x1)

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

s₁ > minus1₁

Status:

minus1₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
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.

(18) QDPOrderProof (EQUIVALENT transformation)

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

The a-transformed usable rules are

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(s(x), 0) → s(x)

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, app(s, x)), app(app(minus, y), x)), 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)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, x), u), z), 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, x2, x3, x4)
s(x1) = s(x1)
minus(x1, x2) = minus(x1)
0 = 0

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

[f1₄, s₁] > [minus₁, 0]

Status:

f1₄: [1,4,3,2]
s₁: [1]
minus₁: [1]
0: []

The following usable rules [FROCOS05] were oriented:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(minus, app(s, x)), 0) → app(s, x)

(19) Obligation:

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

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) 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(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
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.

(23) 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) = APP(x2)
app(x1, x2) = app(x1, x2)
map = map
cons = cons
filter = filter
filter2 = filter2
true = true
false = false
minus = minus
0 = 0
s = s
le = le
if = if
perfectp = perfectp
f = f
nil = nil

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

filter2 > [cons, true] > app₂ > [APP₁, map, filter, minus]
[0, perfectp, f] > le > [cons, true] > app₂ > [APP₁, map, filter, minus]
[0, perfectp, f] > le > [false, s] > app₂ > [APP₁, map, filter, minus]
[0, perfectp, f] > if > [APP₁, map, filter, minus]
nil > [APP₁, map, filter, minus]

Status:

APP₁: [1]
app₂: [2,1]
map: []
cons: []
filter: []
filter2: []
true: []
false: []
minus: []
0: []
s: []
le: []
if: []
perfectp: []
f: []
nil: []

The following usable rules [FROCOS05] were oriented: none

(24) 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(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
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(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
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.

(25) DependencyGraphProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) Overlay + Local Confluence (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) QDPOrderProof (EQUIVALENT transformation)

(9) Obligation:

(10) PisEmptyProof (EQUIVALENT transformation)

(11) TRUE

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) PisEmptyProof (EQUIVALENT transformation)

(16) TRUE

(17) Obligation:

(18) QDPOrderProof (EQUIVALENT transformation)

(19) Obligation:

(20) PisEmptyProof (EQUIVALENT transformation)

(21) TRUE

(22) Obligation:

(23) QDPOrderProof (EQUIVALENT transformation)

(24) Obligation:

(25) DependencyGraphProof (EQUIVALENT transformation)

(26) TRUE