Termination w.r.t. Q proof of /tmp/tmpxttMIz/31.trs

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 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

(0) Obligation:

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

app(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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) 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(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) → APP(:, y)
APP(app(:, app(app(+, x), y)), z) → APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(+, app(app(:, x), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)
APP(app(:, app(app(+, x), y)), z) → APP(:, x)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) → APP(:, y)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(:, app(app(g, z), y)), app(app(+, x), a))
APP(app(:, z), app(app(+, x), app(f, y))) → APP(:, app(app(g, z), y))
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(g, z), y)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(g, z)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(+, x), a)
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(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
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 20 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)

The TRS R consists of the following rules:

app(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
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(app(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)

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)
: = :
+ = +
g = g
a = a

Recursive path order with status [RPO].
Precedence:

app₂ > :
+ > :

Status:

app₂: multiset
:: multiset
+: multiset
g: multiset
a: multiset

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

app(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
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, 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(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
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, 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

Recursive path order with status [RPO].
Precedence:

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

Status:

APP₁: multiset
app₂: multiset
map: multiset
cons: multiset
filter: multiset
filter2: multiset
true: multiset
false: multiset

The following usable rules [FROCOS05] were oriented: none

(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)

The TRS R consists of the following rules:

app(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
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

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: Recursive path order with status [RPO].
Precedence:

cons₂ > map1₂

Status:

map1₂: [2,1]
cons₂: multiset

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(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
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.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) PisEmptyProof (EQUIVALENT transformation)

(9) TRUE

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) PisEmptyProof (EQUIVALENT transformation)

(16) TRUE