Bastard's object
- Caution:TODO: Template:Abbrnot to be confused with: a counter-example, see below
- Caution:
Definition
Examples
Bastard's object Vs a counter-example
Given a statement;
- [ilmath]\forall X[\varphi(X)][/ilmath]^{In words:}^{[Note 1]} made for some [ilmath]\varphi[/ilmath]
to "prove or disprove it" we must establish either the statement holds, or it does not.
Informal discussion:
A counter-example is a demonstration that a statement cannot be true. Suppose for example there is an item, [ilmath]\mathcal{B} [/ilmath] in our language for which [ilmath]\varphi(\mathcal{B})[/ilmath] is known or easily found to be false
This demonstrates a so called "proof by counter-example" as we have given an instance [ilmath]\mathcal{B} [/ilmath] such that we have [ilmath]\neg[\varphi(\mathcal{B})][/ilmath] (the [ilmath]\neg[/ilmath] symbol means "not" or negation, to say [ilmath]\neg[\varphi(\mathcal{B})][/ilmath] holds (is true) means that [ilmath]\varphi(\mathcal{B})[/ilmath] is false, see principle of excluded middle), obviously this means that [ilmath]\varphi(X)[/ilmath] cannot be true for all [ilmath]X[/ilmath]
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:
- [ilmath]\exists Y[\neg[\varphi(Y)]][/ilmath]^{[Note 2]}
Then such a [ilmath]Y[/ilmath] 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
- ↑
Discussion on the basics of reading FOL statements:
This statement, [ilmath]\forall X[\varphi(X)][/ilmath], 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:
- [ilmath]\forall X[\ldots][/ilmath] means "for all things (in our language) it follows that we have ( [ilmath]\ldots[/ilmath] )
- [ilmath]\varphi(X)[/ilmath] is a formula (sometimes overlaps with and is called a predicate)
Combining these:
- [ilmath]\forall X[\varphi(X)][/ilmath] means "for all X there is, the statement [ilmath]\varphi(X)[/ilmath] holds"
- This statement alone doesn't say what [ilmath]\varphi[/ilmath] is, [ilmath]\varphi[/ilmath] 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:
- [ilmath]\forall X[\varphi(X)][/ilmath] is just short for [ilmath]\forall \varphi[\forall X[\varphi(X)]][/ilmath]; note that this statement does not have [ilmath]\varphi[/ilmath] "free" any more.
- [ilmath]\exists\varphi[\forall X[\varphi(X)]][/ilmath] does not have [ilmath]\varphi[/ilmath] free either and says "there exists a [ilmath]\varphi[/ilmath] such that ( for all [ilmath]X[/ilmath] we have ( [ilmath]\varphi(X)[/ilmath] ) )
These are clearly different statements
TODO: Fix formatting - ↑ There exists a "[ilmath]Y[/ilmath]" such that we have ( not the following ( [ilmath]\varphi(Y)[/ilmath] ) )