Difference between revisions of "Bastard's object"

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It would be good to have a category of bastards' objects and a stricter definition. This will do for now. Alec (talk) 11:54, 5 May 2018 (UTC)

Caution:

TODO: Template:Abbr

not to be confused with: a counter-example, see below

Definition

Examples

Trigger's broom

Bastard's object Vs a counter-example

Given a statement;

$\forall X[\varphi(X)]$ ^{In words:}^{[Note 1]} made for some $\varphi$

to "prove or disprove it" we must establish either the statement holds, or it does not.

[Expand]

Informal discussion:

A counter-example is a demonstration that a statement cannot be true. Suppose for example there is an item, $\mathcal{B}$ in our language for which $\varphi(\mathcal{B})$ is known or easily found to be false

This demonstrates a so called "proof by counter-example" as we have given an instance $\mathcal{B}$ such that we have $\neg[\varphi(\mathcal{B})]$ (the $\neg$ symbol means "not" or negation, to say $\neg[\varphi(\mathcal{B})]$ holds (is true) means that $\varphi(\mathcal{B})$ is false, see principle of excluded middle), obviously this means that $\varphi(X)$ cannot be true for all $X$

Informal examples:

the claim "all multiplies of 5 are odd" can instantly be shot down by pointing out that "10 is a multiple of 5 and even"
all sequels are worse than the originals

If we show that:

$\exists Y[\neg[\varphi(Y)]]$ ^{[Note 2]}

Then such a $Y$ is said to be a "counter example".

As such we see that counter examples are useful tools when forming proofs, where as a bastard's object is a concept which shows that something isn't very useful (in terms of logic) or that a concept is informal.

Notes

Jump up ↑
[Expand]
Discussion on the basics of reading FOL statements:
This statement, $\forall X[\varphi(X)]$ , is a first order logic statement (FOL) expressed in a first order language whose symbols are that of logic. See Reading FOL statements for a broader overview. The statement above has 2 parts:
- $\forall X[\ldots]$ means "for all things (in our language) it follows that we have ( $\ldots$ )
- $\varphi(X)$ is a formula (sometimes overlaps with and is called a predicate)
Combining these:
- $\forall X[\varphi(X)]$ means "for all X there is, the statement $\varphi(X)$ holds"
  This statement alone doesn't say what $\varphi$ is, $\varphi$ is said to be a free variable
Any free variables left when one has read an entire statement are assumed to be "for all of them that there are", for example:
- $\forall X[\varphi(X)]$ is just short for $\forall \varphi[\forall X[\varphi(X)]]$ ; note that this statement does not have $\varphi$ "free" any more.
- $\exists\varphi[\forall X[\varphi(X)]]$ does not have $\varphi$ free either and says "there exists a $\varphi$ such that ( for all $X$ we have ( $\varphi(X)$ ) )
These are clearly different statements
TODO: Fix formatting
Jump up ↑ There exists a " $Y$ " such that we have ( not the following ( $\varphi(Y)$ ) )

[FOLinWords1-1] Jump up ↑
[Expand]
Discussion on the basics of reading FOL statements:

This statement, $\forall X[\varphi(X)]$ , is a first order logic statement (FOL) expressed in a first order language whose symbols are that of logic. See Reading FOL statements for a broader overview. The statement above has 2 parts:

$\forall X[\ldots]$ means "for all things (in our language) it follows that we have ( $\ldots$ )

$\varphi(X)$ is a formula (sometimes overlaps with and is called a predicate)

Combining these:

$\forall X[\varphi(X)]$ means "for all X there is, the statement $\varphi(X)$ holds"
This statement alone doesn't say what $\varphi$ is, $\varphi$ is said to be a free variable

Any free variables left when one has read an entire statement are assumed to be "for all of them that there are", for example:

$\forall X[\varphi(X)]$ is just short for $\forall \varphi[\forall X[\varphi(X)]]$ ; note that this statement does not have $\varphi$ "free" any more.

$\exists\varphi[\forall X[\varphi(X)]]$ does not have $\varphi$ free either and says "there exists a $\varphi$ such that ( for all $X$ we have ( $\varphi(X)$ ) )

These are clearly different statements

TODO: Fix formatting

[2] $\forall X[\ldots]$ means "for all things (in our language) it follows that we have ( $\ldots$ )

[3] $\varphi(X)$ is a formula (sometimes overlaps with and is called a predicate)

[4] $\forall X[\varphi(X)]$ means "for all X there is, the statement $\varphi(X)$ holds"
This statement alone doesn't say what $\varphi$ is, $\varphi$ is said to be a free variable

[5] This statement alone doesn't say what $\varphi$ is, $\varphi$ is said to be a free variable

[6] $\forall X[\varphi(X)]$ is just short for $\forall \varphi[\forall X[\varphi(X)]]$ ; note that this statement does not have $\varphi$ "free" any more.

[7] $\exists\varphi[\forall X[\varphi(X)]]$ does not have $\varphi$ free either and says "there exists a $\varphi$ such that ( for all $X$ we have ( $\varphi(X)$ ) )

[2] Jump up ↑ There exists a " $Y$ " such that we have ( not the following ( $\varphi(Y)$ ) )

[Note 1]

[Note 2]

@@ Line 6: / Line 6: @@
 __TOC__
 ==Definition==
+==Examples==
-==={{anchor|VsCounterExample}}Vs a [[counter-example]]===
+* [[Trigger's broom]]
+=={{anchor|VsCounterExample}}Bastard's object Vs a [[counter-example]]==
 Given a statement;
 * {{M|\forall X[\varphi(X)]}}<sup>In words:</sup><ref group="Note" name="FOLinWords1"/> made for some {{M|\varphi}}

Difference between revisions of "Bastard's object"

Latest revision as of 22:15, 8 May 2018

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Bastard's object Vs a counter-example

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