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 QDPSizeChangeProof (⇔)
    ↳7 TRUE
  ↳8 QDP
    ↳9 QDPSizeChangeProof (⇔)
    ↳10 TRUE
  ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 DependencyGraphProof (⇔)
    ↳15 TRUE

(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) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • PART(a_4, l_3) → MATCH_1(a_4, l_3, l_3)
    The graph contains the following edges 1 >= 1, 2 >= 2, 2 >= 3

  • MATCH_1(a_4, l_3, Cons(x, l')) → PART(a_4, l')
    The graph contains the following edges 1 >= 1, 3 > 2

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • APPEND(l1_2, l2_1) → MATCH_0(l1_2, l2_1, l1_2)
    The graph contains the following edges 1 >= 1, 2 >= 2, 1 >= 3

  • MATCH_0(l1_2, l2_1, Cons(x, l)) → APPEND(l, l2_1)
    The graph contains the following edges 3 > 1, 2 >= 2

(10) TRUE

(11) 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.

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MATCH_4(l_5, Cons(a, l')) → MATCH_5(a, l', l_5, part(a, l'))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(Cons(x1, x2)) = 1 + x2   
POL(False) = 0   
POL(MATCH_4(x1, x2)) = x2   
POL(MATCH_5(x1, x2, x3, x4)) = x4   
POL(Nil) = 0   
POL(Pair(x1, x2)) = x1 + x2   
POL(QUICK(x1)) = x1   
POL(True) = 0   
POL(match_1(x1, x2, x3)) = x3   
POL(match_2(x1, x2, x3, x4, x5)) = 1 + x5   
POL(match_3(x1, x2, x3, x4, x5, x6, x7)) = 1 + x1 + x2   
POL(part(x1, x2)) = x2   
POL(test(x1, x2)) = 0   

The following usable rules [FROCOS05] were oriented:

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_1(a_4, l_3, Cons(x, l')) → match_2(x, l', a_4, l_3, part(a_4, l'))
match_3(l1, l2, x, l', a_4, l_3, True) → Pair(l1, Cons(x, l2))
match_3(l1, l2, x, l', a_4, l_3, False) → Pair(Cons(x, l1), l2)
part(a_4, l_3) → match_1(a_4, l_3, l_3)
match_1(a_4, l_3, Nil) → Pair(Nil, Nil)

(13) 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_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.

(14) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(15) TRUE