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Topic: Surprise at my failure to resolve an issue in an elementary paper by Rado
Replies: 44   Last Post: Nov 10, 2013 12:23 PM

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 fom Posts: 1,968 Registered: 12/4/12
Re: Surprise at my failure to resolve an issue in an elementary paper
by Rado

Posted: Nov 3, 2013 11:43 PM
 Plain Text Reply

On 11/3/2013 9:09 PM, fom wrote:
> On 11/3/2013 8:26 PM, David Hartley wrote:
>

When negating the property definition
certain issues raised earlier in the
other thread become apparent

> >
>> But he presumably intended to define L as the
>> set of numbers rho < r which have the property
>>
>> there exist two ordered sets of r numbers differ only at the position
>> indexed by rho which do not have the same image under f.
>>

>
> And, we define sets how?
>
> { x | P(x) }
>
>
> Paul is correct about the quantifier and wrong about failing
> to interpret "whenever ... then" as a conditional.
>
> I see that you also used an existential quantifier in your
> first assessment of the definition.
>
>
> P(k) <-> (
>
> k in { 1, ..., r }
>
> /\
>
> AmAn(
>
> [ ( ( < m > = < a_1, ..., a_r > /\ < n > = < b_1, ..., b_r > )
>
> /\
>
> ( m subset B' /\ n subset B' ) )
>
> /\
>
> Ai( ~( i = k) <-> ( a_i = b_i ) )
>
> /\
>
> ( a_k = b_k ) )
>
> ->
>
> f( m ) =/= f( n ) ]
>
> ) )
>
>
>
>
>

First, as noted in the other thread, definitions
often carry an implicit biconditional. The negation
of the defining property does not make sense without
it. I formulated the above definition with the
conditional to be faithful to the actual words in
the proof.

Second, as noted in the other thread, the restriction
to a single differing index appears to be too strong.
In the definition above, I used

Ai( ~( i = k) <-> ( a_i = b_i ) )

since everyone appears to read the statement that way.

It's negation is

Ei( ( ( i = k) /\ ( a_i = b_i ) ) \/ ( ~( i = k) /\ ~( a_i = b_i ) ) )

Compare this with

Ai( ~( i = k) -> ( a_i = b_i ) )

having negation,

Ei( ~( i = k) /\ ~( a_i = b_i ) )

This is a better choice in the full context of
the negated property in comparison with the
definition of being L-canonical over B.

The definition with only a conditional
fails because negations of formulas such
as

AmAn( p -> q )

proceed as

~Am~( ~An~( ~( p -> q ) ) )

EmEn( p /\ ~q )

You can see that this is a problem with:

Negation of predicate with conditional
======================================

~P(k) <-> (

~( k in { 1, ..., r } )

\/

EmEn(

[ ( ( ( < m > = < a_1, ..., a_r > /\ < n > = < b_1, ..., b_r > )

/\

( m subset B' /\ n subset B' ) )

/\

Ai( ~( i = k) <-> ( a_i = b_i ) )

/\

( a_k = b_k ) )

/\

f( m ) =/= f( n ) ]

) )

In contrast, consider the negated property
with both changes,

Alternate negated property with biconditional
==============================================

~P(k) <-> (

~( k in { 1, ..., r } )

\/

EmEn(

[ ( ( ~( < m > = < a_1, ..., a_r > /\ < n > = < b_1, ..., b_r > )

\/

~( m subset B' /\ n subset B' ) )

\/

Ei( ~( i = k) /\ ~( a_i = b_i ) )

\/

~( a_k = b_k ) )

<->

f( m ) =/= f( n ) ]

) )

The key expressions in this alternate definition
are

Ei( ~( i = k) /\ ~( a_i = b_i ) )

\/

~( a_k = b_k ) )

In other words, if k is not in L, then some
index -- not necessarily k -- must force

f( m ) =/= f( n )

To see why this is the intended interpretation,
look at the biconditional in the definition for
being L-canonical over B from the beginning of
the paper.

Defintion at beginning of paper:

Let L subset {1, 2, ..., r} be given.

Let A and f:[A]^r :=> F be given

LCB(x) denotes "x is L-canonical on B"

===================================================

LCB(x) <-> (

B subset A

/\

AmAn(

( ( < m > = < a_1, ..., a_r > /\ < n > = < b_1, ..., b_r > )

/\

( m subset B /\ n subset B ) )

->

( f( m ) = f( n ) <-> Ak( k in L /\ a_k = b_k ) )

) )

Date Subject Author
11/3/13 Paul
11/3/13 David Hartley
11/3/13 fom
11/3/13 fom
11/3/13 fom
11/4/13 fom
11/4/13 Paul
11/4/13 Paul
11/4/13 Peter Percival
11/4/13 David Hartley
11/4/13 Paul
11/4/13 David Hartley
11/4/13 Paul
11/4/13 David Hartley
11/4/13 Paul
11/5/13 Paul
11/5/13 David Hartley
11/5/13 Paul
11/5/13 David Hartley
11/5/13 Paul
11/6/13 Paul
11/6/13 Paul
11/7/13 Paul
11/7/13 David Hartley
11/7/13 Paul
11/7/13 David Hartley
11/7/13 Paul
11/7/13 David Hartley
11/7/13 David Hartley
11/7/13 Paul
11/7/13 David Hartley
11/8/13 Paul
11/8/13 David Hartley
11/7/13 Paul
11/7/13 fom
11/8/13 Paul
11/8/13 David Hartley
11/10/13 Paul
11/10/13 David Hartley
11/10/13 Paul
11/10/13 David Hartley
11/10/13 David Hartley
11/10/13 Paul
11/4/13 Paul
11/4/13 Peter Percival

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