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As an example of what I'm talking about, the axiom of extension $({\forall}A{\forall}B{\forall}x)(x{\in}A{\iff}x{\in} B){\iff}A=B$ is often used as the definition of set equality.

Why is defining $a$ as being the same as $b$ using $a{\iff}b$ justified?

My doubts stem from things like in the following example:

Let $a$ and $b$ be WFFs.

$a{\iff}b$ means the same as "$a{\implies}b$ and $b{\implies}a$"

$a{\implies}b$ is defined as ${\lnot}(a{\land}{\lnot}b)$

Let $c$ be $x{\land}y$.

Let $d$ be $x{\lor}y$.

Let both $c$ and $d$ be true.

By the above definition of $a{\implies}b$ it's always true that $c{\implies}d$.

Since $c$ is true and $d$ is true, it's also true that ${\lnot}(d{\land}{\lnot}c)$. Therefore $d{\implies}c$

Therefore $c{\iff}d$

However, we know that $x{\land}y$ and $x{\lor}y$ are different things, so it's not true that $c{\iff}d$.

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    $\begingroup$ You assumed c and d were both True. Then the conclusion $c\Leftrightarrow d$ followed from that assumption. $\endgroup$ – Gregory Simon Apr 11 '16 at 15:11
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"$a\iff b$" does not in itself mean that it is a definition.

However, it can be used to express a definition, if we explicitly say that what we're doing is defining $b$ (or $a$). In that case, once we're using that definiton $a\iff b$ will always be true because it's been declared to be the definition.

Note, though, that you can't just pick to random wffs and define them to mean the same. In the particular case of your extensionality axiom however, one of the formulas was $A=B$, which is a previously undefined predicate applied to different variable letters. In that particular case, all that is needed to define a meaning for $=$ is exactly an equivalence of this kind, which will allow you rewrite any atomic formula of the form $t_1=t_2$ into one that doesn't involve the $=$ symbol.

Not all of the communicative content of mathematics is visible in the symbolic formulas. The text around them connects the formulas is also important, oftentimes more than the formulas.


In your particular example, confusion results because you're not careful about whether the reasoning you're doing is supposed to be valid under all interpretations of the variables, or just for a particular one. Here you're explicitly saying, "Let both $c$ and $d$ be true", and everything you conclude thereafter is under that assumption.

It is indeed the case that if both $x\land y$ and $x\lor y$ are true then $x\land y\iff x\lor y$ is also true. That does not imply anything about how $x\land y$ and $x\lor y$ relate to each other in worlds where they are not both true.

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  • $\begingroup$ So you are saying that if we had only the formula of the axiom of extension, we wouldn't know that $(x{\in}A{\iff}x{\in}B)$ means that $A=B$? $\endgroup$ – dsddsadasdasdasdsadasdgfdf Apr 11 '16 at 15:16
  • $\begingroup$ @dsddsadasdasdasdsadasdgfdf It doesn't mean it, it is merely equivalent. The meaning of equality is already known before the Axiom of Extension $\endgroup$ – Hagen von Eitzen Apr 11 '16 at 15:18
  • $\begingroup$ So the Axiom of Extension doesn't define set equality? $\endgroup$ – dsddsadasdasdasdsadasdgfdf Apr 11 '16 at 15:25
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    $\begingroup$ @dsdd: That depend on the precise formulation of ZFC you're using. In some texts, equality is a primitive concept and the Axiom of Extensionality states that sets that have the same elements are equal. Other authors prefer to make $A=B$ a defined notion which abbreviates $\forall x(x\in A\Leftrightarrow x\in B)$ (and this definition will usually not be counted as a named axiom) and then have a separate axiom guaranteeing that if $x=y$ then $x\in A\Leftrightarrow y\in A$ for all $A$. $\endgroup$ – hmakholm left over Monica Apr 11 '16 at 15:57
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You are saying:

Let both $c$ and $d$ be true.

But this is not the only case; we have also the other cases; in particular, when $x$ is true and $y$ is false, we have that:

$x \lor y$ is true and $x \land y$ is false.

Thus:

$x \lor y \Rightarrow x \land y$, i.e. $d \Rightarrow c$ is false.


Regarding Extensionality we have two possibilities; either:

(i) develop set theory using first-order language with equality; in this case the equality is "already there" in the language, and the axiom will be:

$∀x∀y \ [∀z(z \in x ↔ z \in y) → x=y]$;

or

(ii) develop the theory in a language without equality, and introduce the equality with a definition:

$∀x∀y \ [x=y ↔ ∀z(z \in x ↔ z \in y)]$.

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  • $\begingroup$ How are the definitions in $(i)$ and $(ii)$ different? $\endgroup$ – dsddsadasdasdasdsadasdgfdf Apr 11 '16 at 15:47
  • $\begingroup$ @dsddsadasdasdasdsadasdgfdf - from a "practical" point of view, there are no differences. From a "formal" point of view, the first one is not a definition; the symbol $=$ is a primitive symbol of the language and we cannot re-define it. What the axiom add is a way to "prove" the equality between two sets: if we can prove that two sets $a$ and $b$ have all and only the same elements, then the axiom licenses us to assert that they are equal. $\endgroup$ – Mauro ALLEGRANZA Apr 11 '16 at 15:50
  • $\begingroup$ How are we sure that $(ii)$ is actually a definition? As in my original post, just showing that $a{\iff}b$ isn't enough to define $a$ or $b$. $\endgroup$ – dsddsadasdasdasdsadasdgfdf Apr 11 '16 at 15:57

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