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
Usable Rules (Innermost)
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules


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




As we are in the innermost case, we can delete all 15 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
           →DP Problem 4
Size-Change Principle
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules


Dependency Pairs:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. MATCH0(l12, l21, Cons(x, l)) -> APPEND(l, l21)
  2. APPEND(l12, l21) -> MATCH0(l12, l21, l12)
and get the following Size-Change Graph(s):
{1} , {1}
2=2
3>1
{2} , {2}
1=1
1=3
2=2

which lead(s) to this/these maximal multigraph(s):
{1} , {2}
2=2
3>1
3>3
{2} , {1}
1>1
2=2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
Cons(x1, x2) -> Cons(x1, x2)

We obtain no new DP problems.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
Usable Rules (Innermost)
       →DP Problem 3
UsableRules


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




As we are in the innermost case, we can delete all 15 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
           →DP Problem 5
Size-Change Principle
       →DP Problem 3
UsableRules


Dependency Pairs:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. MATCH1(a4, l3, Cons(x, l')) -> PART(a4, l')
  2. PART(a4, l3) -> MATCH1(a4, l3, l3)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
3>2
{2} , {2}
1=1
2=2
2=3

which lead(s) to this/these maximal multigraph(s):
{1} , {2}
1=1
3>2
3>3
{2} , {1}
1=1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
Cons(x1, x2) -> Cons(x1, x2)

We obtain no new DP problems.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
Usable Rules (Innermost)


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




As we are in the innermost case, we can delete all 7 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Negative Polynomial Order


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) -> False
test(x0, y) -> True
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
part(a4, l3) -> match1(a4, l3, l3)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)


Strategy:

innermost




The following Dependency Pairs can be strictly oriented using the given order.

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


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

test(x0, y) -> False
test(x0, y) -> True
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
part(a4, l3) -> match1(a4, l3, l3)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)


Used ordering:
Polynomial Order with Interpretation:

POL( MATCH5(x1, ..., x4) ) = x4

POL( Pair(x1, x2) ) = x1 + x2 + 1

POL( QUICK(x1) ) = x1

POL( MATCH4(x1, x2) ) = x2

POL( Cons(x1, x2) ) = x2 + 1

POL( part(x1, x2) ) = x2 + 1

POL( test(x1, x2) ) = 1

POL( False ) = 1

POL( True ) = 1

POL( match1(x1, ..., x3) ) = x3 + 1

POL( match2(x1, ..., x5) ) = x5 + 1

POL( Nil ) = 0

POL( match3(x1, ..., x7) ) = x1 + x2 + x7 + 1


This results in one new DP problem.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Neg POLO
             ...
               →DP Problem 7
Dependency Graph


Dependency Pairs:

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


Rules:


test(x0, y) -> False
test(x0, y) -> True
match1(a4, l3, Cons(x, l')) -> match2(x, l', a4, l3, part(a4, l'))
match1(a4, l3, Nil) -> Pair(Nil, Nil)
match2(x, l', a4, l3, Pair(l1, l2)) -> match3(l1, l2, x, l', a4, l3, test(a4, x))
part(a4, l3) -> match1(a4, l3, l3)
match3(l1, l2, x, l', a4, l3, True) -> Pair(l1, Cons(x, l2))
match3(l1, l2, x, l', a4, l3, False) -> Pair(Cons(x, l1), l2)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes