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tech / sci.math / Quantifying Conditionals and Conjunctions

SubjectAuthor
* Quantifying Conditionals and ConjunctionsDan Christensen
+* Re: Quantifying Conditionals and ConjunctionsDan Christensen
|`- Re: Quantifying Conditionals and ConjunctionsDan Christensen
`* Re: Quantifying Conditionals and ConjunctionsForest Vaccaro
 `- Re: Quantifying Conditionals and ConjunctionsDan Christensen

1
Quantifying Conditionals and Conjunctions

<44847495-451d-40d1-870e-694505814f05n@googlegroups.com>

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Subject: Quantifying Conditionals and Conjunctions
From: Dan_Chri...@sympatico.ca (Dan Christensen)
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 by: Dan Christensen - Tue, 22 Nov 2022 21:35 UTC

In mathematics, quantifiers are usually restricted to some set. For set S and proposition Q, we can have:

EXAMPLES

(1) ALL(a):[a in S => Q]

(2) EXIST(a):[a in S & Q]

As here, universal quantifiers are usually applied to conditionals ('=>' statements), and existential quantifiers are usually applied to conjunctions ('&' statements). There is a good reason for that:

(3) ALL(a):[a in S & Q] is always FALSE regardless of the truth value of Q

(4) EXIST(a):[a in S => Q] is always TRUE regardless of the truth value of Q

This is probably NOT what you want, though the Drinkers' Paradox is an entertaining variation of (4). (See my posting: https://dcproof.wordpress.com/2014/06/03/the-drinkers-paradox/ )

Proof of (3): ALL(a):[a in S & Q] is always FALSE

From Russell's Paradox of the Universal Set:

1 ALL(a):[Set(a) => EXIST(b):~b in a]
Axiom

Let s be a set

2 Set(s)
Axiom

Apply the above result

3 Set(s) => EXIST(b):~b in s
U Spec, 1

4 EXIST(b):~b in s
Detach, 3, 2

5 ~x in s
E Spec, 4

6 ~x in s | ~Q
Arb Or, 5

7 EXIST(a):[~a in s | ~Q]
E Gen, 6

8 ~ALL(a):~[~a in s | ~Q]
Quant, 7

9 ~ALL(a):~~[~~a in s & ~~Q]
DeMorgan, 8

10 ~ALL(a):[~~a in s & ~~Q]
Rem DNeg, 9

11 ~ALL(a):[a in s & ~~Q]
Rem DNeg, 10

As required:

12 ~ALL(a):[a in s & Q]
Rem DNeg, 11

Proof of (4): EXIST(a):[a in S => Q] is always TRUE

From Russell's Paradox of the Universal Set:

1 ALL(a):[Set(a) => EXIST(b):~b in a]
Axiom

Let s be a set

2 Set(s)
Axiom

Apply the above result

3 Set(s) => EXIST(b):~b in s
U Spec, 1

4 EXIST(b):~b in s
Detach, 3, 2

5 ~x in s
E Spec, 4

6 ~x in s | Q
Arb Or, 5

7 ~~x in s => Q
Imply-Or, 6

8 x in s => Q
Rem DNeg, 7

As required:

9 EXIST(a):[a in s => Q]
E Gen, 8

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Re: Quantifying Conditionals and Conjunctions

<a74d0e63-e218-4609-bada-ea8448a51e8cn@googlegroups.com>

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Subject: Re: Quantifying Conditionals and Conjunctions
From: Dan_Chri...@sympatico.ca (Dan Christensen)
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 by: Dan Christensen - Wed, 23 Nov 2022 02:39 UTC

On Tuesday, November 22, 2022 at 4:35:28 PM UTC-5, Dan Christensen wrote:
> In mathematics, quantifiers are usually restricted to some set. For set S and proposition Q, we can have:
>
> EXAMPLES
>
> (1) ALL(a):[a in S => Q]
>
> (2) EXIST(a):[a in S & Q]
>
> As here, universal quantifiers are usually applied to conditionals ('=>' statements), and existential quantifiers are usually applied to conjunctions ('&' statements). There is a good reason for that:
>
> (3) ALL(a):[a in S & Q] is always FALSE regardless of the truth value of Q
>
> (4) EXIST(a):[a in S => Q] is always TRUE regardless of the truth value of Q
>
> This is probably NOT what you want, though the Drinkers' Paradox is an entertaining variation of (4). (See my posting: https://dcproof.wordpress.com/2014/06/03/the-drinkers-paradox/ )
>
> Proof of (3): ALL(a):[a in S & Q] is always FALSE
>
> From Russell's Paradox of the Universal Set:
>
> 1 ALL(a):[Set(a) => EXIST(b):~b in a]
> Axiom
>
> Let s be a set
>
> 2 Set(s)
> Axiom
>
> Apply the above result
>
> 3 Set(s) => EXIST(b):~b in s
> U Spec, 1
>
> 4 EXIST(b):~b in s
> Detach, 3, 2
>
> 5 ~x in s
> E Spec, 4
>
> 6 ~x in s | ~Q
> Arb Or, 5
>
> 7 EXIST(a):[~a in s | ~Q]
> E Gen, 6
>
> 8 ~ALL(a):~[~a in s | ~Q]
> Quant, 7
>
> 9 ~ALL(a):~~[~~a in s & ~~Q]
> DeMorgan, 8
>
> 10 ~ALL(a):[~~a in s & ~~Q]
> Rem DNeg, 9
>
> 11 ~ALL(a):[a in s & ~~Q]
> Rem DNeg, 10
>
> As required:
>
> 12 ~ALL(a):[a in s & Q]
> Rem DNeg, 11
>
>
> Proof of (4): EXIST(a):[a in S => Q] is always TRUE
>
> From Russell's Paradox of the Universal Set:
>
> 1 ALL(a):[Set(a) => EXIST(b):~b in a]
> Axiom
>
> Let s be a set
>
> 2 Set(s)
> Axiom
>
> Apply the above result
>
> 3 Set(s) => EXIST(b):~b in s
> U Spec, 1
>
> 4 EXIST(b):~b in s
> Detach, 3, 2
>
> 5 ~x in s
> E Spec, 4
>
> 6 ~x in s | Q
> Arb Or, 5
>
> 7 ~~x in s => Q
> Imply-Or, 6
>
> 8 x in s => Q
> Rem DNeg, 7
>
> As required:
>
> 9 EXIST(a):[a in s => Q]
> E Gen, 8
>

By assuming EXIST(a):~P(a), we can prove the predicate logic equivalent: EXIST(a):~P(a) => EXIST(a):[P(a) => Q]

1. EXIST(a):~P(a)
Premise

2. ~P(x)
E Spec, 1

3. ~P(x) | Q
Arb Or, 2

4. ~~P(x) => Q
Imply-Or, 3

5. P(x) => Q
Rem DNeg, 4

6. EXIST(a):[P(a) => Q]
E Gen, 5

7. EXIST(a):~P(a) => EXIST(a):[P(a) => Q]
Conclusion, 1

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Re: Quantifying Conditionals and Conjunctions

<5c0444c1-a2ef-4e65-bee1-3a580c856560n@googlegroups.com>

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Subject: Re: Quantifying Conditionals and Conjunctions
From: Dan_Chri...@sympatico.ca (Dan Christensen)
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 by: Dan Christensen - Wed, 23 Nov 2022 18:56 UTC

On Tuesday, November 22, 2022 at 9:39:28 PM UTC-5, Dan Christensen wrote:
> On Tuesday, November 22, 2022 at 4:35:28 PM UTC-5, Dan Christensen wrote:
> > In mathematics, quantifiers are usually restricted to some set. For set S and proposition Q, we can have:
> >
> > EXAMPLES
> >
> > (1) ALL(a):[a in S => Q]
> >
> > (2) EXIST(a):[a in S & Q]
> >
> > As here, universal quantifiers are usually applied to conditionals ('=>' statements), and existential quantifiers are usually applied to conjunctions ('&' statements). There is a good reason for that:
> >
> > (3) ALL(a):[a in S & Q] is always FALSE regardless of the truth value of Q
> >
> > (4) EXIST(a):[a in S => Q] is always TRUE regardless of the truth value of Q
> >
> > This is probably NOT what you want, though the Drinkers' Paradox is an entertaining variation of (4). (See my posting: https://dcproof.wordpress.com/2014/06/03/the-drinkers-paradox/ )
> >
> > Proof of (3): ALL(a):[a in S & Q] is always FALSE
> >
> > From Russell's Paradox of the Universal Set:
> >
> > 1 ALL(a):[Set(a) => EXIST(b):~b in a]
> > Axiom
> >
> > Let s be a set
> >
> > 2 Set(s)
> > Axiom
> >
> > Apply the above result
> >
> > 3 Set(s) => EXIST(b):~b in s
> > U Spec, 1
> >
> > 4 EXIST(b):~b in s
> > Detach, 3, 2
> >
> > 5 ~x in s
> > E Spec, 4
> >
> > 6 ~x in s | ~Q
> > Arb Or, 5
> >
> > 7 EXIST(a):[~a in s | ~Q]
> > E Gen, 6
> >
> > 8 ~ALL(a):~[~a in s | ~Q]
> > Quant, 7
> >
> > 9 ~ALL(a):~~[~~a in s & ~~Q]
> > DeMorgan, 8
> >
> > 10 ~ALL(a):[~~a in s & ~~Q]
> > Rem DNeg, 9
> >
> > 11 ~ALL(a):[a in s & ~~Q]
> > Rem DNeg, 10
> >
> > As required:
> >
> > 12 ~ALL(a):[a in s & Q]
> > Rem DNeg, 11
> >
> >
> > Proof of (4): EXIST(a):[a in S => Q] is always TRUE
> >
> > From Russell's Paradox of the Universal Set:
> >
> > 1 ALL(a):[Set(a) => EXIST(b):~b in a]
> > Axiom
> >
> > Let s be a set
> >
> > 2 Set(s)
> > Axiom
> >
> > Apply the above result
> >
> > 3 Set(s) => EXIST(b):~b in s
> > U Spec, 1
> >
> > 4 EXIST(b):~b in s
> > Detach, 3, 2
> >
> > 5 ~x in s
> > E Spec, 4
> >
> > 6 ~x in s | Q
> > Arb Or, 5
> >
> > 7 ~~x in s => Q
> > Imply-Or, 6
> >
> > 8 x in s => Q
> > Rem DNeg, 7
> >
> > As required:
> >
> > 9 EXIST(a):[a in s => Q]
> > E Gen, 8
> >
> By assuming EXIST(a):~P(a), we can prove the predicate logic equivalent: EXIST(a):~P(a) => EXIST(a):[P(a) => Q]
>
> 1. EXIST(a):~P(a)
> Premise
>
> 2. ~P(x)
> E Spec, 1
>
> 3. ~P(x) | Q
> Arb Or, 2
>
> 4. ~~P(x) => Q
> Imply-Or, 3
>
> 5. P(x) => Q
> Rem DNeg, 4
>
> 6. EXIST(a):[P(a) => Q]
> E Gen, 5
>
> 7. EXIST(a):~P(a) => EXIST(a):[P(a) => Q]
> Conclusion, 1

By assuming EXIST(a):~P(a), we can prove the predicate logic equivalent: EXIST(a):~P(a) => ~ALL(a):[P(a) & Q(a)] `

1. EXIST(a):~P(a)
Premise

2. ALL(a):[P(a) & Q(a)]
Premise

3. ~P(x)
E Spec, 1

4. P(x) & Q(x)
U Spec, 2

5. P(x)
Split, 4

6. P(x) & ~P(x)
Join, 5, 3

7. ~ALL(a):[P(a) & Q(a)]
Conclusion, 2

8. EXIST(a):~P(a) => ~ALL(a):[P(a) & Q(a)]
Conclusion, 1

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Re: Quantifying Conditionals and Conjunctions

<tllrj0$e83b$1@dont-email.me>

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https://www.novabbs.com/tech/article-flat.php?id=119847&group=sci.math#119847

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From: aso...@ctrsreca.vr (Forest Vaccaro)
Newsgroups: sci.physics.relativity,sci.physics,sci.math
Subject: Re: Quantifying Conditionals and Conjunctions
Date: Wed, 23 Nov 2022 19:16:48 -0000 (UTC)
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 by: Forest Vaccaro - Wed, 23 Nov 2022 19:16 UTC

Dan Christensen wrote:

> In mathematics, quantifiers are usually restricted to some set. For set
> S and proposition Q, we can have: EXAMPLES (1) ALL(a):[a in S => Q]
> (2) EXIST(a):[a in S & Q]

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Re: Quantifying Conditionals and Conjunctions

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 by: Dan Christensen - Wed, 23 Nov 2022 20:39 UTC

On Wednesday, November 23, 2022 at 2:16:56 PM UTC-5, Forest Vaccaro wrote:
> Dan Christensen wrote:
>
> > In mathematics, quantifiers are usually restricted to some set. For set
> > S and proposition Q, we can have: EXAMPLES (1) ALL(a):[a in S => Q]
> > (2) EXIST(a):[a in S & Q]
> you stupid white...

So, what is your Final Solution, naZee Boy? How is Putin's genocidal war of conquest and terror working out for you? I'd go into hiding and stay away from the internet if I were you. Your Red Army ain't what it used to be. XAXAXA

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