/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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WORST_CASE(Omega(n^1), O(n^1))
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 9 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)), 60 ms]
    (8) CdtProblem
    (9) CdtKnowledgeProof [FINISHED, 0 ms]
    (10) BOUNDS(1, 1)
(11) RenamingProof [BOTH BOUNDS(ID, ID), 0 ms]
(12) CpxTRS
    (13) TypeInferenceProof [BOTH BOUNDS(ID, ID), 0 ms]
    (14) typed CpxTrs
    (15) OrderProof [LOWER BOUND(ID), 0 ms]
    (16) typed CpxTrs
    (17) RewriteLemmaProof [LOWER BOUND(ID), 402 ms]
    (18) BEST
        (19) proven lower bound
            (20) LowerBoundPropagationProof [FINISHED, 0 ms]
            (21) BOUNDS(n^1, INF)
        (22) typed CpxTrs


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


The TRS R consists of the following rules:

   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))

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

Rules:
   div(0, z0) -> 0
   div(z0, z1) -> quot(z0, z1, z1)
   quot(0, s(z0), z1) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(div(z0, s(z1)))
Tuples:
   DIV(0, z0) -> c
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(0, s(z0), z1) -> c2
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
S tuples:
   DIV(0, z0) -> c
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(0, s(z0), z1) -> c2
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
K tuples:none
Defined Rule Symbols:   div_2, quot_3

Defined Pair Symbols:   DIV_2, QUOT_3

Compound Symbols:   c, c1_1, c2, c3_1, c4_1


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 2 trailing nodes:
   QUOT(0, s(z0), z1) -> c2
   DIV(0, z0) -> c

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   div(0, z0) -> 0
   div(z0, z1) -> quot(z0, z1, z1)
   quot(0, s(z0), z1) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(div(z0, s(z1)))
Tuples:
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
S tuples:
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
K tuples:none
Defined Rule Symbols:   div_2, quot_3

Defined Pair Symbols:   DIV_2, QUOT_3

Compound Symbols:   c1_1, c3_1, c4_1


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(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   div(0, z0) -> 0
   div(z0, z1) -> quot(z0, z1, z1)
   quot(0, s(z0), z1) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(div(z0, s(z1)))

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
S tuples:
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   DIV_2, QUOT_3

Compound Symbols:   c1_1, c3_1, c4_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.
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(DIV(x_1, x_2)) = x_1 + x_2
   POL(QUOT(x_1, x_2, x_3)) = x_1 + x_3
   POL(c1(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(c4(x_1)) = x_1
   POL(s(x_1)) = [1] + x_1

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
S tuples:
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
K tuples:
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:   DIV_2, QUOT_3

Compound Symbols:   c1_1, c3_1, c4_1


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(9) CdtKnowledgeProof (FINISHED)
The following tuples could be moved from S to K by knowledge propagation:
   QUOT(z0, 0, s(z1)) -> c4(DIV(z0, s(z1)))
   DIV(z0, z1) -> c1(QUOT(z0, z1, z1))
   QUOT(s(z0), s(z1), z2) -> c3(QUOT(z0, z1, z2))
Now S is empty
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(10)
BOUNDS(1, 1)

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(11) RenamingProof (BOTH BOUNDS(ID, ID))
Renamed function symbols to avoid clashes with predefined symbol.
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(12)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   div(0', y) -> 0'
   div(x, y) -> quot(x, y, y)
   quot(0', s(y), z) -> 0'
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0', s(z)) -> s(div(x, s(z)))

S is empty.
Rewrite Strategy: INNERMOST
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(13) TypeInferenceProof (BOTH BOUNDS(ID, ID))
Infered types.
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(14)
Obligation:
Innermost TRS:
Rules:
div(0', y) -> 0'
div(x, y) -> quot(x, y, y)
quot(0', s(y), z) -> 0'
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0', s(z)) -> s(div(x, s(z)))

Types:
div :: 0':s -> 0':s -> 0':s
0' :: 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
s :: 0':s -> 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat -> 0':s

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(15) OrderProof (LOWER BOUND(ID))
Heuristically decided to analyse the following defined symbols:
div, quot

They will be analysed ascendingly in the following order:
div = quot

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(16)
Obligation:
Innermost TRS:
Rules:
div(0', y) -> 0'
div(x, y) -> quot(x, y, y)
quot(0', s(y), z) -> 0'
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0', s(z)) -> s(div(x, s(z)))

Types:
div :: 0':s -> 0':s -> 0':s
0' :: 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
s :: 0':s -> 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat -> 0':s


Generator Equations:
gen_0':s2_0(0) <=> 0'
gen_0':s2_0(+(x, 1)) <=> s(gen_0':s2_0(x))


The following defined symbols remain to be analysed:
quot, div

They will be analysed ascendingly in the following order:
div = quot

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(17) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
quot(gen_0':s2_0(n4_0), gen_0':s2_0(+(1, n4_0)), gen_0':s2_0(c)) -> gen_0':s2_0(0), rt in Omega(1 + n4_0)

Induction Base:
quot(gen_0':s2_0(0), gen_0':s2_0(+(1, 0)), gen_0':s2_0(c)) ->_R^Omega(1)
0'

Induction Step:
quot(gen_0':s2_0(+(n4_0, 1)), gen_0':s2_0(+(1, +(n4_0, 1))), gen_0':s2_0(c)) ->_R^Omega(1)
quot(gen_0':s2_0(n4_0), gen_0':s2_0(+(1, n4_0)), gen_0':s2_0(c)) ->_IH
gen_0':s2_0(0)

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
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(18)
Complex Obligation (BEST)

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(19)
Obligation:
Proved the lower bound n^1 for the following obligation:

Innermost TRS:
Rules:
div(0', y) -> 0'
div(x, y) -> quot(x, y, y)
quot(0', s(y), z) -> 0'
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0', s(z)) -> s(div(x, s(z)))

Types:
div :: 0':s -> 0':s -> 0':s
0' :: 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
s :: 0':s -> 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat -> 0':s


Generator Equations:
gen_0':s2_0(0) <=> 0'
gen_0':s2_0(+(x, 1)) <=> s(gen_0':s2_0(x))


The following defined symbols remain to be analysed:
quot, div

They will be analysed ascendingly in the following order:
div = quot

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(20) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(21)
BOUNDS(n^1, INF)

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(22)
Obligation:
Innermost TRS:
Rules:
div(0', y) -> 0'
div(x, y) -> quot(x, y, y)
quot(0', s(y), z) -> 0'
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0', s(z)) -> s(div(x, s(z)))

Types:
div :: 0':s -> 0':s -> 0':s
0' :: 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
s :: 0':s -> 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat -> 0':s


Lemmas:
quot(gen_0':s2_0(n4_0), gen_0':s2_0(+(1, n4_0)), gen_0':s2_0(c)) -> gen_0':s2_0(0), rt in Omega(1 + n4_0)


Generator Equations:
gen_0':s2_0(0) <=> 0'
gen_0':s2_0(+(x, 1)) <=> s(gen_0':s2_0(x))


The following defined symbols remain to be analysed:
div

They will be analysed ascendingly in the following order:
div = quot