Termination w.r.t. Q proof of /tmp/tmp7IwdCb/08.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 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

(0) Obligation:

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

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
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(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
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(D, app(app(+, x), y)) → APP(app(+, app(D, x)), app(D, y))
APP(D, app(app(+, x), y)) → APP(+, app(D, x))
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(*, x), y)) → APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(*, x), y)) → APP(+, app(app(*, y), app(D, x)))
APP(D, app(app(*, x), y)) → APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) → APP(*, y)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(app(-, app(D, x)), app(D, y))
APP(D, app(app(-, x), y)) → APP(-, app(D, x))
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(D, 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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
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(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
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 19 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(D, y)

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
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(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
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

D1(+(x, y)) → D1(y)
D1(+(x, y)) → D1(x)
D1(*(x, y)) → D1(x)
D1(*(x, y)) → D1(y)
D1(-(x, y)) → D1(x)
D1(-(x, y)) → D1(y)

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(D, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

-₂: [1,2]
*₂: [1,2]
D1₁: [1]
+₂: [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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
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(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
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, 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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
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(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
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, 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) = APP(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:

APP₁ > map > filter
app₂ > map > filter
cons > map > filter
filter2 > filter
true > filter
false > filter

Status:

APP₁: [1]
cons: []
true: []
map: []
false: []
app₂: [1,2]
filter2: []
filter: []

The following usable rules [FROCOS05] were oriented: none

(14) 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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
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(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
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) QDPOrderProof (EQUIVALENT transformation)

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

(16) Obligation:

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

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
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(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(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) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) PisEmptyProof (EQUIVALENT transformation)

(18) TRUE