Termination w.r.t. Q proof of /tmp/tmpPUurrs/#3.2.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 DependencyGraphProof (⇔)

    ↳21 TRUE

(0) Obligation:

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

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, 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(app(minus, x), app(s, y)) → APP(pred, app(app(minus, x), y))
APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), y))
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, x)
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(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, 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 3 SCCs with 14 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

The TRS R consists of the following rules:

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, 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

minus1(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, x), app(s, y)) → APP(app(minus, x), y)

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

s₁ > minus1₂

Status:

minus1₂: [1,2]
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(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, 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(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))

The TRS R consists of the following rules:

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, 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)

quot1(s(x), s(y)) → quot1(minus(x, y), s(y))

The a-transformed usable rules are

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
quot1(x1, x2) = x1
s(x1) = s(x1)
minus(x1, x2) = x1
0 = 0
pred(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

0 > s₁

Status:

s₁: [1]
0: []

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, 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) 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(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, 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.

(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(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(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, 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
nil = nil
minus = minus
s = s
pred = pred
quot = quot
0 = 0

Lexicographic path order with status [LPO].
Precedence:

map > APP₁ > app₂
cons > APP₁ > app₂
filter > APP₁ > app₂
filter2 > APP₁ > app₂
true > app₂
false > APP₁ > app₂
minus > app₂
s > app₂
quot > app₂

Status:

APP₁: [1]
minus: []
true: []
pred: []
s: []
0: []
filter: []
quot: []
cons: []
map: []
false: []
app₂: [1,2]
filter2: []
nil: []

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

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(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, 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.

(20) 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) DependencyGraphProof (EQUIVALENT transformation)

(21) TRUE