WORST_CASE(?,O(n^2))
proof of input_fKzxHMbrlg.trs
# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished


The Runtime Complexity (parallel-innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(2) CdtProblem
(3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
(4) CdtProblem
(5) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 109 ms]
(6) CdtProblem
(7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 93 ms]
(8) CdtProblem
(9) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
(10) CdtProblem
(11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 50 ms]
(12) CdtProblem
(13) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
(14) BOUNDS(1, 1)


----------------------------------------

(0)
Obligation:
The Runtime Complexity (parallel-innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

   sort(nil) -> nil
   sort(cons(x, y)) -> insert(x, sort(y))
   insert(x, nil) -> cons(x, nil)
   insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
   choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
   choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
   choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)

S is empty.
Rewrite Strategy: PARALLEL_INNERMOST
----------------------------------------

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS with rewrite strategy PARALLEL_INNERMOST to CDT
----------------------------------------

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   sort(nil) -> nil
   sort(cons(z0, z1)) -> insert(z0, sort(z1))
   insert(z0, nil) -> cons(z0, nil)
   insert(z0, cons(z1, z2)) -> choose(z0, cons(z1, z2), z0, z1)
   choose(z0, cons(z1, z2), z3, 0) -> cons(z0, cons(z1, z2))
   choose(z0, cons(z1, z2), 0, s(z3)) -> cons(z1, insert(z0, z2))
   choose(z0, cons(z1, z2), s(z3), s(z4)) -> choose(z0, cons(z1, z2), z3, z4)
Tuples:
   SORT(nil) -> c
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, nil) -> c2
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), z3, 0) -> c4
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:
   SORT(nil) -> c
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, nil) -> c2
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), z3, 0) -> c4
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
K tuples:none
Defined Rule Symbols:   sort_1, insert_2, choose_4

Defined Pair Symbols:   SORT_1, INSERT_2, CHOOSE_4

Compound Symbols:   c, c1_2, c2, c3_1, c4, c5_1, c6_1


----------------------------------------

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 3 trailing nodes:
   INSERT(z0, nil) -> c2
   CHOOSE(z0, cons(z1, z2), z3, 0) -> c4
   SORT(nil) -> c

----------------------------------------

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   sort(nil) -> nil
   sort(cons(z0, z1)) -> insert(z0, sort(z1))
   insert(z0, nil) -> cons(z0, nil)
   insert(z0, cons(z1, z2)) -> choose(z0, cons(z1, z2), z0, z1)
   choose(z0, cons(z1, z2), z3, 0) -> cons(z0, cons(z1, z2))
   choose(z0, cons(z1, z2), 0, s(z3)) -> cons(z1, insert(z0, z2))
   choose(z0, cons(z1, z2), s(z3), s(z4)) -> choose(z0, cons(z1, z2), z3, z4)
Tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
K tuples:none
Defined Rule Symbols:   sort_1, insert_2, choose_4

Defined Pair Symbols:   SORT_1, INSERT_2, CHOOSE_4

Compound Symbols:   c1_2, c3_1, c5_1, c6_1


----------------------------------------

(5) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
We considered the (Usable) Rules:none
And the Tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(CHOOSE(x_1, x_2, x_3, x_4)) = x_1
   POL(INSERT(x_1, x_2)) = x_1
   POL(SORT(x_1)) = x_1
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c3(x_1)) = x_1
   POL(c5(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(choose(x_1, x_2, x_3, x_4)) = [1] + x_1 + x_2
   POL(cons(x_1, x_2)) = [1] + x_1 + x_2
   POL(insert(x_1, x_2)) = [1] + x_1 + x_2
   POL(nil) = [1]
   POL(s(x_1)) = [1] + x_1
   POL(sort(x_1)) = [1] + x_1

----------------------------------------

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   sort(nil) -> nil
   sort(cons(z0, z1)) -> insert(z0, sort(z1))
   insert(z0, nil) -> cons(z0, nil)
   insert(z0, cons(z1, z2)) -> choose(z0, cons(z1, z2), z0, z1)
   choose(z0, cons(z1, z2), z3, 0) -> cons(z0, cons(z1, z2))
   choose(z0, cons(z1, z2), 0, s(z3)) -> cons(z1, insert(z0, z2))
   choose(z0, cons(z1, z2), s(z3), s(z4)) -> choose(z0, cons(z1, z2), z3, z4)
Tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
K tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
Defined Rule Symbols:   sort_1, insert_2, choose_4

Defined Pair Symbols:   SORT_1, INSERT_2, CHOOSE_4

Compound Symbols:   c1_2, c3_1, c5_1, c6_1


----------------------------------------

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
We considered the (Usable) Rules:
   choose(z0, cons(z1, z2), 0, s(z3)) -> cons(z1, insert(z0, z2))
   sort(nil) -> nil
   insert(z0, cons(z1, z2)) -> choose(z0, cons(z1, z2), z0, z1)
   sort(cons(z0, z1)) -> insert(z0, sort(z1))
   insert(z0, nil) -> cons(z0, nil)
   choose(z0, cons(z1, z2), z3, 0) -> cons(z0, cons(z1, z2))
   choose(z0, cons(z1, z2), s(z3), s(z4)) -> choose(z0, cons(z1, z2), z3, z4)
And the Tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(CHOOSE(x_1, x_2, x_3, x_4)) = [2]x_2
   POL(INSERT(x_1, x_2)) = [2]x_2
   POL(SORT(x_1)) = [2]x_1^2
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c3(x_1)) = x_1
   POL(c5(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(choose(x_1, x_2, x_3, x_4)) = [1] + x_2
   POL(cons(x_1, x_2)) = [1] + x_2
   POL(insert(x_1, x_2)) = [1] + x_2
   POL(nil) = 0
   POL(s(x_1)) = 0
   POL(sort(x_1)) = x_1

----------------------------------------

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   sort(nil) -> nil
   sort(cons(z0, z1)) -> insert(z0, sort(z1))
   insert(z0, nil) -> cons(z0, nil)
   insert(z0, cons(z1, z2)) -> choose(z0, cons(z1, z2), z0, z1)
   choose(z0, cons(z1, z2), z3, 0) -> cons(z0, cons(z1, z2))
   choose(z0, cons(z1, z2), 0, s(z3)) -> cons(z1, insert(z0, z2))
   choose(z0, cons(z1, z2), s(z3), s(z4)) -> choose(z0, cons(z1, z2), z3, z4)
Tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
K tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
Defined Rule Symbols:   sort_1, insert_2, choose_4

Defined Pair Symbols:   SORT_1, INSERT_2, CHOOSE_4

Compound Symbols:   c1_2, c3_1, c5_1, c6_1


----------------------------------------

(9) CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
The following tuples could be moved from S to K by knowledge propagation:
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))

----------------------------------------

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   sort(nil) -> nil
   sort(cons(z0, z1)) -> insert(z0, sort(z1))
   insert(z0, nil) -> cons(z0, nil)
   insert(z0, cons(z1, z2)) -> choose(z0, cons(z1, z2), z0, z1)
   choose(z0, cons(z1, z2), z3, 0) -> cons(z0, cons(z1, z2))
   choose(z0, cons(z1, z2), 0, s(z3)) -> cons(z1, insert(z0, z2))
   choose(z0, cons(z1, z2), s(z3), s(z4)) -> choose(z0, cons(z1, z2), z3, z4)
Tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
K tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
Defined Rule Symbols:   sort_1, insert_2, choose_4

Defined Pair Symbols:   SORT_1, INSERT_2, CHOOSE_4

Compound Symbols:   c1_2, c3_1, c5_1, c6_1


----------------------------------------

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
We considered the (Usable) Rules:
   choose(z0, cons(z1, z2), 0, s(z3)) -> cons(z1, insert(z0, z2))
   sort(nil) -> nil
   insert(z0, cons(z1, z2)) -> choose(z0, cons(z1, z2), z0, z1)
   sort(cons(z0, z1)) -> insert(z0, sort(z1))
   insert(z0, nil) -> cons(z0, nil)
   choose(z0, cons(z1, z2), z3, 0) -> cons(z0, cons(z1, z2))
   choose(z0, cons(z1, z2), s(z3), s(z4)) -> choose(z0, cons(z1, z2), z3, z4)
And the Tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(CHOOSE(x_1, x_2, x_3, x_4)) = [2]x_3 + x_1*x_2
   POL(INSERT(x_1, x_2)) = [2]x_1 + x_1*x_2
   POL(SORT(x_1)) = [2]x_1^2
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c3(x_1)) = x_1
   POL(c5(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(choose(x_1, x_2, x_3, x_4)) = [2] + x_1 + x_2
   POL(cons(x_1, x_2)) = [2] + x_1 + x_2
   POL(insert(x_1, x_2)) = [2] + x_1 + x_2
   POL(nil) = 0
   POL(s(x_1)) = [2] + x_1
   POL(sort(x_1)) = [2] + x_1

----------------------------------------

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   sort(nil) -> nil
   sort(cons(z0, z1)) -> insert(z0, sort(z1))
   insert(z0, nil) -> cons(z0, nil)
   insert(z0, cons(z1, z2)) -> choose(z0, cons(z1, z2), z0, z1)
   choose(z0, cons(z1, z2), z3, 0) -> cons(z0, cons(z1, z2))
   choose(z0, cons(z1, z2), 0, s(z3)) -> cons(z1, insert(z0, z2))
   choose(z0, cons(z1, z2), s(z3), s(z4)) -> choose(z0, cons(z1, z2), z3, z4)
Tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:none
K tuples:
   SORT(cons(z0, z1)) -> c1(INSERT(z0, sort(z1)), SORT(z1))
   CHOOSE(z0, cons(z1, z2), 0, s(z3)) -> c5(INSERT(z0, z2))
   INSERT(z0, cons(z1, z2)) -> c3(CHOOSE(z0, cons(z1, z2), z0, z1))
   CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) -> c6(CHOOSE(z0, cons(z1, z2), z3, z4))
Defined Rule Symbols:   sort_1, insert_2, choose_4

Defined Pair Symbols:   SORT_1, INSERT_2, CHOOSE_4

Compound Symbols:   c1_2, c3_1, c5_1, c6_1


----------------------------------------

(13) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
----------------------------------------

(14)
BOUNDS(1, 1)