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Four Letters ROLLO


TheCube
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BLUE

As BIZE=2 then {BI??, B?Z?, B??E, ?IZ?, ?I?E, ??ZE}

As BYTE=2 then {BY??, B?T?, B??E, ?YT?, ?Y?E, ??TE}

Creating a matrix, and cross-referencing each possibility

results in the following: {BYZ?, BIT?, B??E, ?YZE, ?ITE}

There is no valid word BYZ? or ?YZE, thus {BIT?, B??E, ?ITE}.

If BLUE>=2, then B??E is proven

Edited by Dej Mar
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BLUE

NEON - 0  TUCK - 0

CUBE - 1  BIZE - 2

LASS - 0  BYTE - 2

BULB - 1  BUCK - 1

POEM - 0  BITE - 2

Dej Mar - 25, TheCube - 10

SCORES COMING SOON!

PS - @HG I know what you were trying to say but the logic seemed incomplete to me.

EDIT: adding Ps.

Edited by TheCube
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????


Score Changes: none yet

(PS. From earlier proofs given that were accepted, HG should have had +5, and though I don't like giving up points I would accept the loss. I think HG was trying to say was, since tUCK=0, and Bulb, Bize and Byte >=1, with BUCK=1 then B--- is proven.)

Edited by Dej Mar
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???? 

HOPE - 0


Score Changes: none yet

(P.S. to the P.S. (HG's proof did not require "and Bulb, Bize and Byte >=1", that was just added to overflow the proverbial cup. HG only lacked giving "since TUCK=0". If no objection, I will adjust the scores accordingly).

Edited by Dej Mar
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????  

HOPE - 0 

EDIT - 0 

CRAP - 0

ABLE - 0

SPIT - 0


Score Changes: none yet

(no objections noted: Scores adjusted. Dej Mar -5, Hidden G +5

TheCube - 335

Molly Mae - 248

WombatBreath - 227

Fabpig - 162

Hidden G - 142

Dej Mar - 112

kristmark1 - 93

Aaryan - 60

unreality - 37

Bong - 39

mEEster-Michael - 31

sks - 26

Shadow7 - 25

curr3nt - 20

Shakingdavid - 17

Quag - 10

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