Term Rewriting System R:
[x0, y, l12, l21, x, l, a4, l3, l', l1, l2, l5, a]
test(x0, y) -> True
test(x0, y) -> False
append(l12, l21) -> match0(l12, l21, l12)
match0(l12, l21, Nil) -> l21
match0(l12, l21, Cons(x, l)) -> Cons(x, append(l, l21))
part(a4, l3) -> match1(a4, l3, l3)
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
quick(l5) -> match4(l5, l5)
match4(l5, Nil) -> Nil
match4(l5, Cons(a, l')) -> match5(a, l', l5, part(a, l'))
match5(a, l', l5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

APPEND(l12, l21) -> MATCH0(l12, l21, l12)
MATCH0(l12, l21, Cons(x, l)) -> APPEND(l, l21)
PART(a4, l3) -> MATCH1(a4, l3, l3)
MATCH1(a4, l3, Cons(x, l')) -> MATCH2(x, l', a4, l3, part(a4, l'))
MATCH1(a4, l3, Cons(x, l')) -> PART(a4, l')
MATCH2(x, l', a4, l3, Pair(l1, l2)) -> MATCH3(l1, l2, x, l', a4, l3, test(a4, x))
MATCH2(x, l', a4, l3, Pair(l1, l2)) -> TEST(a4, x)
QUICK(l5) -> MATCH4(l5, l5)
MATCH4(l5, Cons(a, l')) -> MATCH5(a, l', l5, part(a, l'))
MATCH4(l5, Cons(a, l')) -> PART(a, l')
MATCH5(a, l', l5, Pair(l1, l2)) -> APPEND(quick(l1), Cons(a, quick(l2)))
MATCH5(a, l', l5, Pair(l1, l2)) -> QUICK(l1)
MATCH5(a, l', l5, Pair(l1, l2)) -> QUICK(l2)

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pairs:

MATCH0(l12, l21, Cons(x, l)) -> APPEND(l, l21)
APPEND(l12, l21) -> MATCH0(l12, l21, l12)

Rules:

test(x0, y) -> True
test(x0, y) -> False
append(l12, l21) -> match0(l12, l21, l12)
match0(l12, l21, Nil) -> l21
match0(l12, l21, Cons(x, l)) -> Cons(x, append(l, l21))
part(a4, l3) -> match1(a4, l3, l3)
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
quick(l5) -> match4(l5, l5)
match4(l5, Nil) -> Nil
match4(l5, Cons(a, l')) -> match5(a, l', l5, part(a, l'))
match5(a, l', l5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MATCH0(l12, l21, Cons(x, l)) -> APPEND(l, l21)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(MATCH_0(x1, x2, x3)) =  x3 POL(Cons(x1, x2)) =  1 + x2 POL(APPEND(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

APPEND(l12, l21) -> MATCH0(l12, l21, l12)

Rules:

test(x0, y) -> True
test(x0, y) -> False
append(l12, l21) -> match0(l12, l21, l12)
match0(l12, l21, Nil) -> l21
match0(l12, l21, Cons(x, l)) -> Cons(x, append(l, l21))
part(a4, l3) -> match1(a4, l3, l3)
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
quick(l5) -> match4(l5, l5)
match4(l5, Nil) -> Nil
match4(l5, Cons(a, l')) -> match5(a, l', l5, part(a, l'))
match5(a, l', l5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pairs:

MATCH1(a4, l3, Cons(x, l')) -> PART(a4, l')
PART(a4, l3) -> MATCH1(a4, l3, l3)

Rules:

test(x0, y) -> True
test(x0, y) -> False
append(l12, l21) -> match0(l12, l21, l12)
match0(l12, l21, Nil) -> l21
match0(l12, l21, Cons(x, l)) -> Cons(x, append(l, l21))
part(a4, l3) -> match1(a4, l3, l3)
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
quick(l5) -> match4(l5, l5)
match4(l5, Nil) -> Nil
match4(l5, Cons(a, l')) -> match5(a, l', l5, part(a, l'))
match5(a, l', l5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MATCH1(a4, l3, Cons(x, l')) -> PART(a4, l')

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(PART(x1, x2)) =  x2 POL(Cons(x1, x2)) =  1 + x2 POL(MATCH_1(x1, x2, x3)) =  x3

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

PART(a4, l3) -> MATCH1(a4, l3, l3)

Rules:

test(x0, y) -> True
test(x0, y) -> False
append(l12, l21) -> match0(l12, l21, l12)
match0(l12, l21, Nil) -> l21
match0(l12, l21, Cons(x, l)) -> Cons(x, append(l, l21))
part(a4, l3) -> match1(a4, l3, l3)
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
quick(l5) -> match4(l5, l5)
match4(l5, Nil) -> Nil
match4(l5, Cons(a, l')) -> match5(a, l', l5, part(a, l'))
match5(a, l', l5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polynomial Ordering`

Dependency Pairs:

MATCH5(a, l', l5, Pair(l1, l2)) -> QUICK(l2)
MATCH5(a, l', l5, Pair(l1, l2)) -> QUICK(l1)
MATCH4(l5, Cons(a, l')) -> MATCH5(a, l', l5, part(a, l'))
QUICK(l5) -> MATCH4(l5, l5)

Rules:

test(x0, y) -> True
test(x0, y) -> False
append(l12, l21) -> match0(l12, l21, l12)
match0(l12, l21, Nil) -> l21
match0(l12, l21, Cons(x, l)) -> Cons(x, append(l, l21))
part(a4, l3) -> match1(a4, l3, l3)
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
quick(l5) -> match4(l5, l5)
match4(l5, Nil) -> Nil
match4(l5, Cons(a, l')) -> match5(a, l', l5, part(a, l'))
match5(a, l', l5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MATCH4(l5, Cons(a, l')) -> MATCH5(a, l', l5, part(a, l'))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

part(a4, l3) -> match1(a4, l3, l3)
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 6`
`             ↳Dependency Graph`

Dependency Pairs:

MATCH5(a, l', l5, Pair(l1, l2)) -> QUICK(l2)
MATCH5(a, l', l5, Pair(l1, l2)) -> QUICK(l1)
QUICK(l5) -> MATCH4(l5, l5)

Rules:

test(x0, y) -> True
test(x0, y) -> False
append(l12, l21) -> match0(l12, l21, l12)
match0(l12, l21, Nil) -> l21
match0(l12, l21, Cons(x, l)) -> Cons(x, append(l, l21))
part(a4, l3) -> match1(a4, l3, l3)
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
quick(l5) -> match4(l5, l5)
match4(l5, Nil) -> Nil
match4(l5, Cons(a, l')) -> match5(a, l', l5, part(a, l'))
match5(a, l', l5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes