Termination w.r.t. Q proof of /tmp/tmpQI11pK/#3.22.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

  ↳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(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), 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) 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(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0))))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, y), app(app(times, app(s, z)), 0))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, app(s, z)), 0)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(times, app(s, z))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))
APP(app(times, x), app(s, y)) → APP(app(plus, app(app(times, x), y)), x)
APP(app(times, x), app(s, y)) → APP(plus, app(app(times, x), y))
APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(plus, x), app(s, y)) → APP(s, app(app(plus, x), y))
APP(app(plus, x), app(s, y)) → APP(app(plus, x), 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(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), 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.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 17 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), 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.
We have to consider all minimal (P,Q,R)-chains.

(6) 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

plus1(x, s(y)) → plus1(x, y)

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

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

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

[plus1₂, s₁]

Status:

plus1₂: [2,1]
s₁: [1]

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(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), 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.
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(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))
APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), 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.
We have to consider all minimal (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

times1(x, plus(y, s(z))) → times1(x, s(z))
times1(x, s(y)) → times1(x, y)
times1(x, plus(y, s(z))) → times1(x, plus(y, times(s(z), 0)))

The a-transformed usable rules are

times(x, 0) → 0
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

The following pairs can be oriented strictly and are deleted.

APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
times1(x1, x2) = times1(x1, x2)
plus(x1, x2) = plus(x1, x2)
s(x1) = x1
times(x1, x2) = times
0 = 0

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

[times1₂, plus₂, times, 0]

Status:

times1₂: [2,1]
plus₂: [2,1]
times: []
0: []

The following usable rules [FROCOS05] were oriented:

app(app(times, x), 0) → 0
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))

(12) Obligation:

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

APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), 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.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

times1(x, s(y)) → times1(x, y)
times1(x, plus(y, s(z))) → times1(x, plus(y, times(s(z), 0)))

The a-transformed usable rules are

times(x, 0) → 0
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

The following pairs can be oriented strictly and are deleted.

APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
times1(x1, x2) = times1(x1, x2)
s(x1) = s(x1)
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times
0 = 0

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

plus₂ > times1₂ > [s₁, times] > 0

Status:

times1₂: [1,2]
s₁: [1]
plus₂: [1,2]
times: []
0: []

The following usable rules [FROCOS05] were oriented:

app(app(times, x), 0) → 0
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))

(14) Obligation:

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

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), 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.
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(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), 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.
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
times = times
plus = plus
s = s
0 = 0
nil = nil

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

cons > [app₂, map, true] > filter2 > APP₁
cons > [app₂, map, true] > 0 > APP₁
[filter, nil] > [app₂, map, true] > filter2 > APP₁
[filter, nil] > [app₂, map, true] > 0 > APP₁
false > APP₁
s > plus > times > [app₂, map, true] > filter2 > APP₁
s > plus > times > [app₂, map, true] > 0 > APP₁

Status:

APP₁: [1]
app₂: [2,1]
map: []
cons: []
filter: []
filter2: []
true: []
false: []
times: []
plus: []
s: []
0: []
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(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), 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.
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) 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

(17) Obligation:

(18) QDPOrderProof (EQUIVALENT transformation)

(19) Obligation:

(20) DependencyGraphProof (EQUIVALENT transformation)

(21) TRUE