Termination w.r.t. Q proof of /tmp/tmp1oHxq0/wiehe03.trs

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

  ↳11 QDP

    ↳12 QDPOrderProof (⇔)

    ↳13 QDP

    ↳14 PisEmptyProof (⇔)

    ↳15 TRUE

  ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 PisEmptyProof (⇔)

    ↳20 TRUE

  ↳21 QDP

(0) Obligation:

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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(cons(x, l), k) → APP(l, k)
SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))
SUM(cons(x, cons(y, l))) → PLUS(x, y)
SUM(app(l, cons(x, cons(y, k)))) → SUM(app(l, sum(cons(x, cons(y, k)))))
SUM(app(l, cons(x, cons(y, k)))) → APP(l, sum(cons(x, cons(y, k))))
SUM(app(l, cons(x, cons(y, k)))) → SUM(cons(x, cons(y, k)))
SUM(plus(cons(0, x), cons(y, l))) → PRED(sum(cons(s(x), cons(y, l))))
SUM(plus(cons(0, x), cons(y, l))) → SUM(cons(s(x), cons(y, l)))
PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
PLUS(s(x), s(y)) → IF(gt(x, y), x, y)
PLUS(s(x), s(y)) → GT(x, y)
PLUS(s(x), s(y)) → IF(not(gt(x, y)), id(x), id(y))
PLUS(s(x), s(y)) → NOT(gt(x, y))
PLUS(s(x), s(y)) → ID(x)
PLUS(s(x), s(y)) → ID(y)
PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
PLUS(s(x), x) → IF(gt(x, x), id(x), id(x))
PLUS(s(x), x) → GT(x, x)
PLUS(s(x), x) → ID(x)
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
PLUS(id(x), s(y)) → IF(gt(s(y), y), y, s(y))
PLUS(id(x), s(y)) → GT(s(y), y)
NOT(x) → IF(x, false, true)
GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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 5 SCCs with 17 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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.

GT(s(x), s(y)) → GT(x, y)

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

Lexicographic Path Order [LPO].
Precedence:

s₁ > GT₁

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(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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:

PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) Obligation:

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

APP(cons(x, l), k) → APP(l, k)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(cons(x, l), k) → APP(l, k)

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

Lexicographic Path Order [LPO].
Precedence:

cons₂ > APP₁

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) Obligation:

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

SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SUM(x1) = x1
cons(x1, x2) = cons(x2)
plus(x1, x2) = x2
s(x1) = s(x1)
if(x1, x2, x3) = if(x1)
gt(x1, x2) = x2
not(x1) = not(x1)
id(x1) = id(x1)
zero = zero
true = true
false = false

Lexicographic Path Order [LPO].
Precedence:

s₁ > if₁
s₁ > id₁
not₁ > if₁

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) Obligation:

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

SUM(app(l, cons(x, cons(y, k)))) → SUM(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.