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Continuation of Free and bound variables in "if" statements

definitions:

A number is even if it is divisible by $2$.

The number is even if it is divisible by $2$.

Is the usage of the and a different in mathematical definitions(sorry for the poor example above, if you have any better example that would be appreciated). Also aren't mathematical definitions assertions ?

A number $n$ is even if it is divisible by $2$.

Every number $n$ is even if it is divisible by $2$.

I know that a definition is NOT an assertion and thus cannot be true or false(Right ?). The definition just describes a property(ies) of some mathematical object.

Now for the above two definitons is $n$ a free variable in the first but bound variable in the second ? Also the second definition can be expressed as $\forall n (n\ is\ even\ \leftrightarrow n\ is\ divisible\ by\ 2)$. Doesn't that mean that a definition in a sense is true or false since it can be expressed by a logical symbols ?

Another definition:

Suppose $n$ is an integer. Then $n$ is even if it is divisible by $2$.

Where is the location of Suppose $n$ is an integer(a precondition) in the logical structure of a definition(after or before the biarrow) ? $\forall n (n\ is\ an\ integer \rightarrow n\ is\ even\ \leftrightarrow n\ is\ divisible\ by\ 2)$ or $\forall n (n\ is\ an\ integer \land n\ is\ even\ \leftrightarrow n\ is\ divisible\ by\ 2)$ or $\forall n ( n\ is\ even\ \leftrightarrow n\ is\ an\ integer \land n\ is\ divisible\ by\ 2)$

Also when trying to prove "$x$ is even" what exactly should I do ?

Another definition:

Suppose $R$ is a partial order on a set $A$, $B ⊆ A$, and $ b ∈ B$. Then $ b$ is called an $R$-smallest element of B (or just a smallest element if $R$ is clear from the context) if $∀x ∈ B(bRx)$.

How do I express this definiton in logical symbols(are $A$,$b$,$B$, $R$ free or bound?)?What should I do if I want to prove that $z$ is a $H$ smallest element of $M$ ?

Suppose $f : A → B$ and $C ⊆ A$. The set $f ∩ (C × B)$, which is a relation from $C$ to $B$, is called the restriction of f to C, and is sometimes denoted $f|C$. In other words, $f|C = f∩(C × B)$.

What is the location of $f|C$ in the logical structure ?

Also, it would be great to give a list of books that could clarify misconceptions of these type.

Nameless
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1 Answers1

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Some theory

See George Tourlakis, Lectures in Logic and Set Theory. Volume 1 : Mathematical Logic (2003), I.7. Defined Symbols, page 112-on :

We start with a theory $\Gamma$, spoken in some basic formal language $L$. As the development of $\Gamma$ proceeds, gradually and continuously we extend $L$ into languages $L_n$, for $n ≥ 0$ (we have set $L_0 = L$). The theory itself is being extended by stages, as a sequence $\Gamma_n$, $n ≥ 0$. [...] At that same stage we also add to $\Gamma_n$ the defining nonlogical axiom of the new symbol in question.

Specifically, if $\varphi(x_1, ..., x_n)$ is some formula, we then can introduce a new predicate symbol “$P$” that stands for $\varphi$.

This entails adding $P$ to $L_k$ as a newn-ary predicate symbol, and adding

$P(x_1, ... x_n) ↔ \varphi(x_1, ..., x_n)$

to $\Gamma_k$ as the defining axiom for $P$.

Thus, we can start with the formula $\varphi(n) := \exists y(n = 2 \times y)$ and extend the "basic" language with the new predicate $Even(n)$ and the theory with the defining axiom :

$Even(n) \leftrightarrow \exists y(n = 2 \times y)$.

This is "implicitly" universally quantified, i.e. : $\forall n [Even(n) \leftrightarrow \exists y(n = 2 \times y) ]$.

Comment

About the natural language, my "feeling" is that :

a number $n$ is even if it is divisible by $2$

is a "correct" form for a definition; "a number $n$" must be interpreted as "a number $n$ whatver" i.e. as having $n$ universally quantified.

I would prefer to "read" :

every number $n$ is even if it is divisible by $2$

as meaning : "for every number $n$, $n$ is even if it is divisible by $2$", which amount to the same statement.

Question (1) :

when trying to prove "$x$ is even" what exactly should I do ?

$Even(n)$ is defined as : $\exists y(n = 2 \times y)$; thus, proving that e.g. $6 = 2 \times 3$, by rule of logic we can derive : $\exists y(6 = 2 \times y)$, which - by definition - is :

$Even(6)$.

Question (2) :

suppose $R$ is a partial order on a set $A, B⊆A$, and $b∈B$. Then $b$ is called an $R$-smallest element of $B$ (or just a smallest element if $R$ is clear from the context) if $∀x∈B(bRx)$. How do I express this definiton in logical symbols(are $A,b,B,R$ free or bound?)?

See Tourlakis, cit, page 114 :

In this case we have to add a new $n$-ary function symbol $f$ into $L_k$ by a definition. That is, we add $f$ to $L_k$ and also add the following formula to $\Gamma_k$ as a new nonlogical axiom:

$y = f(y_1, ... y_n) ↔ \varphi(y, y_1, ... y_n)$

provided we have a proof in $\Gamma_k$ of the formula : $(∃!y) \varphi(y, y_1, ... y_n)$.

In our example, assuming for simplicity that we have already introduced with a definition the symbol $PO(x,y)$ as an abbreviation for "$x$ is a Partial Order on $y$", we can introduce the following symbol $min(x,y,z)$ as an abbreviation for "the smallest element of $z$ for the relation $x$ on $y$" by the definition :

$\forall R \forall A \forall B \forall b \quad [ b = min(R,A,B) \leftrightarrow (R \subseteq A \times A \land PO(R,A) \land B \subseteq A \land b \in B \land \forall x (x\in B \rightarrow bRx)) ]$

Now, according to this definition, the answer to the question :

what should I do if I want to prove that $z$ is a $H$-smallest element of $M$ ?

is straightforward. You have to prove that, for $z \in M$ :

$\forall x (x\in M \rightarrow zHx)$.

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Mauro ALLEGRANZA
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  • Great Answer but here is a couple of points I don't understand: 1) $R⊆A×A∧PO(R,A)∧B⊆A∧b∈B→∀x(x∈B→bRx)$ OR $(R⊆A×A∧PO(R,A)∧B⊆A∧b∈B)→∀x(x∈B→bRx)$ 2)why $→∀x(x∈B→bRx)$ not $\leftrightarrow ∀x(x∈B→bRx)$. I thought if in definitions meant if and only if – Nameless Jun 05 '14 at 13:06
  • @Nameless - you are right for 1) : it is better to group between parentheses all the conjuncts [but note that $(A \land B) \rightarrow C$ is equivalent to : $A \rightarrow (B \rightarrow C)$]. Regarding 2) : NO, the $\leftrightarrow$ of the definition is that between the definiendum (i.e.$b=min(R,A,B)$) and the definiens (i.e.the "big" sub-formula on the right). – Mauro ALLEGRANZA Jun 05 '14 at 13:14
  • Ok. Lets say we have this Definition: Every integer is even if it is divisible by 2. is same as If n is an integer then n is even if it is divisible by 2. means $\forall n (Even(n)\leftrightarrow $ (n is integer $\land \exists k(n=2k)))$ (right ?) is it the same as $\forall n (Even(n)\leftrightarrow $ (n is integer $\rightarrow \exists k(n=2k)))$ ?? – Nameless Jun 05 '14 at 13:28
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    Also why $→∀x(x∈B→bRx) $ and not $\land ∀x(x∈B→bRx)$. Just to clarify it: you put the precondition before the $\rightarrow$ because these are always given, because without them the definition will not apply to the object we want. And the thing we want to prove after the $\rightarrow$. In this case this ∀n(Even(n)↔ (n is integer →∃k(n=2k))) is right while the other one wrong(is this correct ?). Right ? – Nameless Jun 05 '14 at 13:34
  • Why did you change ∀x(x∈B→bRx) to ∀x(x∈B∧bRx). The first one is right when we have a universal quantifier ?
    • – Nameless Jun 05 '14 at 14:31
  • So if have to prove that z is a H -smallest element of M ? what should I do now ?? Should I prove $R⊆A×A∧PO(R,A)∧B⊆A∧b∈B∧∀x(x∈B∧bRx)$ or $∀x(x∈M∧zHx)$. The thing that is annoying me is that if we don't have $R⊆A×A∧PO(R,A)∧B⊆A∧b∈B$ as given(aka suppose) then the definition doesn't even make sense(right? we need preconditions as they specify the type of structure the definition applies to) so when proving $b$ is an $R$ -smallest element of $B$ it doesn't make sense to prove $R⊆A×A∧PO(R,A)∧B⊆A∧b∈B$ as they should be given. – Nameless Jun 05 '14 at 14:44
  • But if the logical structure of the definition $[b=min(R,A,B)↔(R⊆A×A∧PO(R,A)∧B⊆A∧b∈B∧∀x(x∈B∧bRx))]$ means we have to prove (R⊆A×A∧PO(R,A)∧B⊆A∧b∈B) is true but aren't these given(as without them the definition doesn't make sense). So in this sense $(R⊆A×A∧PO(R,A)∧B⊆A∧b∈B)→∀x(x∈B→bRx))$ makes better sense since if we want to prove the truth of $(R⊆A×A∧PO(R,A)∧B⊆A∧b∈B)→∀x(x∈B→bRx))$ we assume $(R⊆A×A∧PO(R,A)∧B⊆A∧b∈B)$ as given. Do you get my confusion here ? Just so you know I take the definition as a biconditional assertion that is always true. – Nameless Jun 05 '14 at 14:48
  • YES. If $R$ is not a relation or $B$ is not a subset of $A$ or ... it makes no sense to ask if $b \in B$ is an $R$-smallest element of $B$. Thus, applying the def, if $R$ is not a relation on $A$, we have that the long RHS will be false and so, by property of equivalence, will be false also $b=min(R,A,B)$ and we can agree that in this case it is not true to say that $b$ is the $R$-smallest element of $B$. – Mauro ALLEGRANZA Jun 05 '14 at 14:53
  • Also I think the $∀x(x∈B→bRx)$ should be used with the assertion that $B\ne ∅$ on the RHS as using $∀x(x∈B \land bRx)$ is always false($\forall x(x>0)$) since (for example) potato(or any random real number) is not in $B$. – Nameless Jun 05 '14 at 14:58
  • STOP I've restored the original version, because the condition $b \in B$ is enough for $B \ne \emptyset$. – Mauro ALLEGRANZA Jun 05 '14 at 15:05
  • Yeah...hahahaha. I didn't notice that $b\in B$ makes $B\ne ∅$ . So just to make sure I get this "idea" perfectly: if I want to prove that ** $b$ is an $R $-smallest element of $B$** I have to "technically" prove $(R⊆A×A∧PO(R,A)∧B⊆A∧b∈B∧∀x(x∈B→bRx))$ but this part $(R⊆A×A∧PO(R,A)∧B⊆A∧b∈B)$ is always given to me in the "Suppose" part of the question because if I don't have this $(R⊆A×A∧PO(R,A)∧B⊆A∧b∈B)$ than the question is meaningless as the LHS will be automatically false. Did I get it right ? – Nameless Jun 05 '14 at 15:14
  • @Nameless - YES. – Mauro ALLEGRANZA Jun 05 '14 at 15:34
  • THANK YOU A MILLION(I finally understand it)... I deeply appreciate your help. – Nameless Jun 05 '14 at 15:48