WORST_CASE(?,O(n^1))
proof of input_v2CzHZxIBr.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# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished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^1).

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(2) CdtProblem
(3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
(4) CdtProblem
(5) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
(6) CdtProblem
(7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 111 ms]
(8) CdtProblem
(9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 28 ms]
(10) CdtProblem
(11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 23 ms]
(12) CdtProblem
(13) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
(14) BOUNDS(1, 1)


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


The TRS R consists of the following rules:

   msort(nil) -> nil
   msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, nil) -> nil
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

S is empty.
Rewrite Strategy: PARALLEL_INNERMOST
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(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS with rewrite strategy PARALLEL_INNERMOST to CDT
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(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   msort(nil) -> nil
   msort(.(z0, z1)) -> .(min(z0, z1), msort(del(min(z0, z1), .(z0, z1))))
   min(z0, nil) -> z0
   min(z0, .(z1, z2)) -> if(<=(z0, z1), min(z0, z2), min(z1, z2))
   del(z0, nil) -> nil
   del(z0, .(z1, z2)) -> if(=(z0, z1), z2, .(z1, del(z0, z2)))
Tuples:
   MSORT(nil) -> c
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, nil) -> c3
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, nil) -> c6
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
S tuples:
   MSORT(nil) -> c
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, nil) -> c3
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, nil) -> c6
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
K tuples:none
Defined Rule Symbols:   msort_1, min_2, del_2

Defined Pair Symbols:   MSORT_1, MIN_2, DEL_2

Compound Symbols:   c, c1_1, c2_3, c3, c4_1, c5_1, c6, c7_1


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 3 trailing nodes:
   DEL(z0, nil) -> c6
   MIN(z0, nil) -> c3
   MSORT(nil) -> c

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

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   msort(nil) -> nil
   msort(.(z0, z1)) -> .(min(z0, z1), msort(del(min(z0, z1), .(z0, z1))))
   min(z0, nil) -> z0
   min(z0, .(z1, z2)) -> if(<=(z0, z1), min(z0, z2), min(z1, z2))
   del(z0, nil) -> nil
   del(z0, .(z1, z2)) -> if(=(z0, z1), z2, .(z1, del(z0, z2)))
Tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
S tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
K tuples:none
Defined Rule Symbols:   msort_1, min_2, del_2

Defined Pair Symbols:   MSORT_1, MIN_2, DEL_2

Compound Symbols:   c1_1, c2_3, c4_1, c5_1, c7_1


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(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   msort(nil) -> nil
   msort(.(z0, z1)) -> .(min(z0, z1), msort(del(min(z0, z1), .(z0, z1))))

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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   del(z0, .(z1, z2)) -> if(=(z0, z1), z2, .(z1, del(z0, z2)))
   del(z0, nil) -> nil
   min(z0, nil) -> z0
   min(z0, .(z1, z2)) -> if(<=(z0, z1), min(z0, z2), min(z1, z2))
Tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
S tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
K tuples:none
Defined Rule Symbols:   del_2, min_2

Defined Pair Symbols:   MSORT_1, MIN_2, DEL_2

Compound Symbols:   c1_1, c2_3, c4_1, c5_1, c7_1


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(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(.(x_1, x_2)) = [1] + x_1 + x_2
   POL(<=(x_1, x_2)) = [1] + x_1
   POL(=(x_1, x_2)) = 0
   POL(DEL(x_1, x_2)) = 0
   POL(MIN(x_1, x_2)) = 0
   POL(MSORT(x_1)) = [1]
   POL(c1(x_1)) = x_1
   POL(c2(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c4(x_1)) = x_1
   POL(c5(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(del(x_1, x_2)) = [1] + x_1 + x_2
   POL(if(x_1, x_2, x_3)) = x_1 + x_3
   POL(min(x_1, x_2)) = [1] + x_1 + x_2
   POL(nil) = [1]

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   del(z0, .(z1, z2)) -> if(=(z0, z1), z2, .(z1, del(z0, z2)))
   del(z0, nil) -> nil
   min(z0, nil) -> z0
   min(z0, .(z1, z2)) -> if(<=(z0, z1), min(z0, z2), min(z1, z2))
Tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
S tuples:
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
K tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
Defined Rule Symbols:   del_2, min_2

Defined Pair Symbols:   MSORT_1, MIN_2, DEL_2

Compound Symbols:   c1_1, c2_3, c4_1, c5_1, c7_1


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(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
We considered the (Usable) Rules:
   del(z0, .(z1, z2)) -> if(=(z0, z1), z2, .(z1, del(z0, z2)))
And the Tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(.(x_1, x_2)) = [1] + x_1 + x_2
   POL(<=(x_1, x_2)) = [1] + x_1 + x_2
   POL(=(x_1, x_2)) = 0
   POL(DEL(x_1, x_2)) = 0
   POL(MIN(x_1, x_2)) = x_1 + x_2
   POL(MSORT(x_1)) = [1] + x_1
   POL(c1(x_1)) = x_1
   POL(c2(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c4(x_1)) = x_1
   POL(c5(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(del(x_1, x_2)) = 0
   POL(if(x_1, x_2, x_3)) = x_1
   POL(min(x_1, x_2)) = [1] + x_1 + x_2
   POL(nil) = 0

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(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   del(z0, .(z1, z2)) -> if(=(z0, z1), z2, .(z1, del(z0, z2)))
   del(z0, nil) -> nil
   min(z0, nil) -> z0
   min(z0, .(z1, z2)) -> if(<=(z0, z1), min(z0, z2), min(z1, z2))
Tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
S tuples:
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
K tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
Defined Rule Symbols:   del_2, min_2

Defined Pair Symbols:   MSORT_1, MIN_2, DEL_2

Compound Symbols:   c1_1, c2_3, c4_1, c5_1, c7_1


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(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
We considered the (Usable) Rules:
   del(z0, .(z1, z2)) -> if(=(z0, z1), z2, .(z1, del(z0, z2)))
And the Tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(.(x_1, x_2)) = [1] + x_1 + x_2
   POL(<=(x_1, x_2)) = [1] + x_1 + x_2
   POL(=(x_1, x_2)) = 0
   POL(DEL(x_1, x_2)) = x_2
   POL(MIN(x_1, x_2)) = 0
   POL(MSORT(x_1)) = x_1
   POL(c1(x_1)) = x_1
   POL(c2(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c4(x_1)) = x_1
   POL(c5(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(del(x_1, x_2)) = 0
   POL(if(x_1, x_2, x_3)) = x_1
   POL(min(x_1, x_2)) = [1] + x_1 + x_2
   POL(nil) = 0

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

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   del(z0, .(z1, z2)) -> if(=(z0, z1), z2, .(z1, del(z0, z2)))
   del(z0, nil) -> nil
   min(z0, nil) -> z0
   min(z0, .(z1, z2)) -> if(<=(z0, z1), min(z0, z2), min(z1, z2))
Tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
S tuples:none
K tuples:
   MSORT(.(z0, z1)) -> c1(MIN(z0, z1))
   MSORT(.(z0, z1)) -> c2(MSORT(del(min(z0, z1), .(z0, z1))), DEL(min(z0, z1), .(z0, z1)), MIN(z0, z1))
   MIN(z0, .(z1, z2)) -> c4(MIN(z0, z2))
   MIN(z0, .(z1, z2)) -> c5(MIN(z1, z2))
   DEL(z0, .(z1, z2)) -> c7(DEL(z0, z2))
Defined Rule Symbols:   del_2, min_2

Defined Pair Symbols:   MSORT_1, MIN_2, DEL_2

Compound Symbols:   c1_1, c2_3, c4_1, c5_1, c7_1


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(13) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
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(14)
BOUNDS(1, 1)