Given that W is trivial (courtesy of Andy Latto), KNOW = NO establishes the triviality of K. CELL = SELL ==> C = S SCENT = CENT ==> S is trivial (and therefore so is C) ... Rarer letters are harder to destroy, but: CUE = QUEUE ==> UE = QUEUE ==> QUE is trivial We've already shown that E is an identity, so QU is trivial. U can be destroyed by FLOWER = FLOUR ==> WE = U, so Q is also trivial. At the moment, we're left with: _B_D_FG_IJKLM__P_R___V_XYZ KNIGHT = NIGHT easily gets rid of K, and we have Y = I = E (from TYRE = TIRE and COMPLIMENT = COMPLEMENT), so those all go: _B_D_FG__JKLM__P_R___V_X_Z GNOME = NOME (from elliptic functions) kills the G. _B_D_F___JKLM__P_R___V_X_Z Also, FAT = PHAT to deduce F = P, but I can't immediately trivialise either.
Sent: Friday, September 25, 2015 at 9:32 PM From: "Andy Latto" <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] something linguistic
And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically,
WHETHER = WEATHER shows that A is trivial
WHICH = WITCH shows that T is trivial.
DAMN = DAM, so N is trivial
On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
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