(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) 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.
(7) 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.