Termination w.r.t. Q proof of /tmp/tmpYHV0r

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 DependencyGraphProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 DependencyGraphProof (⇔)

    ↳14 TRUE

  ↳15 QDP

(0) Obligation:

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

test(x_0, y) → True
test(x_0, y) → False
append(l1_2, l2_1) → match_0(l1_2, l2_1, l1_2)
match_0(l1_2, l2_1, Nil) → l2_1
match_0(l1_2, l2_1, Cons(x, l)) → Cons(x, append(l, l2_1))
part(a_4, l_3) → match_1(a_4, l_3, l_3)
match_1(a_4, l_3, Nil) → Pair(Nil, Nil)
match_1(a_4, l_3, Cons(x, l')) → match_2(x, l', a_4, l_3, part(a_4, l'))
match_2(x, l', a_4, l_3, Pair(l1, l2)) → match_3(l1, l2, x, l', a_4, l_3, test(a_4, x))
match_3(l1, l2, x, l', a_4, l_3, False) → Pair(Cons(x, l1), l2)
match_3(l1, l2, x, l', a_4, l_3, True) → Pair(l1, Cons(x, l2))
quick(l_5) → match_4(l_5, l_5)
match_4(l_5, Nil) → Nil
match_4(l_5, Cons(a, l')) → match_5(a, l', l_5, part(a, l'))
match_5(a, l', l_5, Pair(l1, l2)) → append(quick(l1), Cons(a, quick(l2)))

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:

APPEND(l1_2, l2_1) → MATCH_0(l1_2, l2_1, l1_2)
MATCH_0(l1_2, l2_1, Cons(x, l)) → APPEND(l, l2_1)
PART(a_4, l_3) → MATCH_1(a_4, l_3, l_3)
MATCH_1(a_4, l_3, Cons(x, l')) → MATCH_2(x, l', a_4, l_3, part(a_4, l'))
MATCH_1(a_4, l_3, Cons(x, l')) → PART(a_4, l')
MATCH_2(x, l', a_4, l_3, Pair(l1, l2)) → MATCH_3(l1, l2, x, l', a_4, l_3, test(a_4, x))
MATCH_2(x, l', a_4, l_3, Pair(l1, l2)) → TEST(a_4, x)
QUICK(l_5) → MATCH_4(l_5, l_5)
MATCH_4(l_5, Cons(a, l')) → MATCH_5(a, l', l_5, part(a, l'))
MATCH_4(l_5, Cons(a, l')) → PART(a, l')
MATCH_5(a, l', l_5, Pair(l1, l2)) → APPEND(quick(l1), Cons(a, quick(l2)))
MATCH_5(a, l', l_5, Pair(l1, l2)) → QUICK(l1)
MATCH_5(a, l', l_5, Pair(l1, l2)) → QUICK(l2)

The TRS R consists of the following rules:

test(x_0, y) → True
test(x_0, y) → False
append(l1_2, l2_1) → match_0(l1_2, l2_1, l1_2)
match_0(l1_2, l2_1, Nil) → l2_1
match_0(l1_2, l2_1, Cons(x, l)) → Cons(x, append(l, l2_1))
part(a_4, l_3) → match_1(a_4, l_3, l_3)
match_1(a_4, l_3, Nil) → Pair(Nil, Nil)
match_1(a_4, l_3, Cons(x, l')) → match_2(x, l', a_4, l_3, part(a_4, l'))
match_2(x, l', a_4, l_3, Pair(l1, l2)) → match_3(l1, l2, x, l', a_4, l_3, test(a_4, x))
match_3(l1, l2, x, l', a_4, l_3, False) → Pair(Cons(x, l1), l2)
match_3(l1, l2, x, l', a_4, l_3, True) → Pair(l1, Cons(x, l2))
quick(l_5) → match_4(l_5, l_5)
match_4(l_5, Nil) → Nil
match_4(l_5, Cons(a, l')) → match_5(a, l', l_5, part(a, l'))
match_5(a, l', l_5, Pair(l1, l2)) → append(quick(l1), Cons(a, quick(l2)))

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 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

MATCH_1(a_4, l_3, Cons(x, l')) → PART(a_4, l')
PART(a_4, l_3) → MATCH_1(a_4, l_3, l_3)

The TRS R consists of the following rules:

test(x_0, y) → True
test(x_0, y) → False
append(l1_2, l2_1) → match_0(l1_2, l2_1, l1_2)
match_0(l1_2, l2_1, Nil) → l2_1
match_0(l1_2, l2_1, Cons(x, l)) → Cons(x, append(l, l2_1))
part(a_4, l_3) → match_1(a_4, l_3, l_3)
match_1(a_4, l_3, Nil) → Pair(Nil, Nil)
match_1(a_4, l_3, Cons(x, l')) → match_2(x, l', a_4, l_3, part(a_4, l'))
match_2(x, l', a_4, l_3, Pair(l1, l2)) → match_3(l1, l2, x, l', a_4, l_3, test(a_4, x))
match_3(l1, l2, x, l', a_4, l_3, False) → Pair(Cons(x, l1), l2)
match_3(l1, l2, x, l', a_4, l_3, True) → Pair(l1, Cons(x, l2))
quick(l_5) → match_4(l_5, l_5)
match_4(l_5, Nil) → Nil
match_4(l_5, Cons(a, l')) → match_5(a, l', l_5, part(a, l'))
match_5(a, l', l_5, Pair(l1, l2)) → append(quick(l1), Cons(a, quick(l2)))

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.

MATCH_1(a_4, l_3, Cons(x, l')) → PART(a_4, l')

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MATCH_1(x1, x2, x3) = x3
Cons(x1, x2) = Cons(x1, x2)
PART(x1, x2) = x2

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

PART(a_4, l_3) → MATCH_1(a_4, l_3, l_3)

The TRS R consists of the following rules:

test(x_0, y) → True
test(x_0, y) → False
append(l1_2, l2_1) → match_0(l1_2, l2_1, l1_2)
match_0(l1_2, l2_1, Nil) → l2_1
match_0(l1_2, l2_1, Cons(x, l)) → Cons(x, append(l, l2_1))
part(a_4, l_3) → match_1(a_4, l_3, l_3)
match_1(a_4, l_3, Nil) → Pair(Nil, Nil)
match_1(a_4, l_3, Cons(x, l')) → match_2(x, l', a_4, l_3, part(a_4, l'))
match_2(x, l', a_4, l_3, Pair(l1, l2)) → match_3(l1, l2, x, l', a_4, l_3, test(a_4, x))
match_3(l1, l2, x, l', a_4, l_3, False) → Pair(Cons(x, l1), l2)
match_3(l1, l2, x, l', a_4, l_3, True) → Pair(l1, Cons(x, l2))
quick(l_5) → match_4(l_5, l_5)
match_4(l_5, Nil) → Nil
match_4(l_5, Cons(a, l')) → match_5(a, l', l_5, part(a, l'))
match_5(a, l', l_5, Pair(l1, l2)) → append(quick(l1), Cons(a, quick(l2)))

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

(8) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(9) TRUE

(10) Obligation:

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

MATCH_0(l1_2, l2_1, Cons(x, l)) → APPEND(l, l2_1)
APPEND(l1_2, l2_1) → MATCH_0(l1_2, l2_1, l1_2)

The TRS R consists of the following rules:

test(x_0, y) → True
test(x_0, y) → False
append(l1_2, l2_1) → match_0(l1_2, l2_1, l1_2)
match_0(l1_2, l2_1, Nil) → l2_1
match_0(l1_2, l2_1, Cons(x, l)) → Cons(x, append(l, l2_1))
part(a_4, l_3) → match_1(a_4, l_3, l_3)
match_1(a_4, l_3, Nil) → Pair(Nil, Nil)
match_1(a_4, l_3, Cons(x, l')) → match_2(x, l', a_4, l_3, part(a_4, l'))
match_2(x, l', a_4, l_3, Pair(l1, l2)) → match_3(l1, l2, x, l', a_4, l_3, test(a_4, x))
match_3(l1, l2, x, l', a_4, l_3, False) → Pair(Cons(x, l1), l2)
match_3(l1, l2, x, l', a_4, l_3, True) → Pair(l1, Cons(x, l2))
quick(l_5) → match_4(l_5, l_5)
match_4(l_5, Nil) → Nil
match_4(l_5, Cons(a, l')) → match_5(a, l', l_5, part(a, l'))
match_5(a, l', l_5, Pair(l1, l2)) → append(quick(l1), Cons(a, quick(l2)))

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.

MATCH_0(l1_2, l2_1, Cons(x, l)) → APPEND(l, l2_1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MATCH_0(x1, x2, x3) = x3
Cons(x1, x2) = Cons(x1, x2)
APPEND(x1, x2) = x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

APPEND(l1_2, l2_1) → MATCH_0(l1_2, l2_1, l1_2)

The TRS R consists of the following rules:

test(x_0, y) → True
test(x_0, y) → False
append(l1_2, l2_1) → match_0(l1_2, l2_1, l1_2)
match_0(l1_2, l2_1, Nil) → l2_1
match_0(l1_2, l2_1, Cons(x, l)) → Cons(x, append(l, l2_1))
part(a_4, l_3) → match_1(a_4, l_3, l_3)
match_1(a_4, l_3, Nil) → Pair(Nil, Nil)
match_1(a_4, l_3, Cons(x, l')) → match_2(x, l', a_4, l_3, part(a_4, l'))
match_2(x, l', a_4, l_3, Pair(l1, l2)) → match_3(l1, l2, x, l', a_4, l_3, test(a_4, x))
match_3(l1, l2, x, l', a_4, l_3, False) → Pair(Cons(x, l1), l2)
match_3(l1, l2, x, l', a_4, l_3, True) → Pair(l1, Cons(x, l2))
quick(l_5) → match_4(l_5, l_5)
match_4(l_5, Nil) → Nil
match_4(l_5, Cons(a, l')) → match_5(a, l', l_5, part(a, l'))
match_5(a, l', l_5, Pair(l1, l2)) → append(quick(l1), Cons(a, quick(l2)))

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

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(14) TRUE

(15) Obligation:

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

MATCH_5(a, l', l_5, Pair(l1, l2)) → QUICK(l1)
QUICK(l_5) → MATCH_4(l_5, l_5)
MATCH_4(l_5, Cons(a, l')) → MATCH_5(a, l', l_5, part(a, l'))
MATCH_5(a, l', l_5, Pair(l1, l2)) → QUICK(l2)

The TRS R consists of the following rules:

test(x_0, y) → True
test(x_0, y) → False
append(l1_2, l2_1) → match_0(l1_2, l2_1, l1_2)
match_0(l1_2, l2_1, Nil) → l2_1
match_0(l1_2, l2_1, Cons(x, l)) → Cons(x, append(l, l2_1))
part(a_4, l_3) → match_1(a_4, l_3, l_3)
match_1(a_4, l_3, Nil) → Pair(Nil, Nil)
match_1(a_4, l_3, Cons(x, l')) → match_2(x, l', a_4, l_3, part(a_4, l'))
match_2(x, l', a_4, l_3, Pair(l1, l2)) → match_3(l1, l2, x, l', a_4, l_3, test(a_4, x))
match_3(l1, l2, x, l', a_4, l_3, False) → Pair(Cons(x, l1), l2)
match_3(l1, l2, x, l', a_4, l_3, True) → Pair(l1, Cons(x, l2))
quick(l_5) → match_4(l_5, l_5)
match_4(l_5, Nil) → Nil
match_4(l_5, Cons(a, l')) → match_5(a, l', l_5, part(a, l'))
match_5(a, l', l_5, Pair(l1, l2)) → append(quick(l1), Cons(a, quick(l2)))

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