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In classical propositional logic, or (aka. disjunction) is a Boolean function $\vee:\{0,1\}^2\longrightarrow\{0,1\}$ defined as $\vee:=\{ (00,0), (01,1), (10,1),(11,1) \}$, and and (aka. conjunction) is a Boolean function $\wedge: \{0,1\}^2\longrightarrow\{0,1\}$ defined as $\{(00,0),(01,0),(10,0),(11,1)\}$.

There's also rules of inference for these functions, such as $\vee$-introduction, $\wedge$-introduction, $\vee$-elimination, and $\wedge$-elimination, as well as other machinery from logic.


So, or and and are not the same object. But when using them in math reasoning, sometimes the 2 can be used interchangeably.

Example. Let $X$ be a topological space, let $A\subseteq X$, let $\text{Lim }A$ be the set of limit points of $A$. The closure of $A$ is defined as $A\cup\text{Lim }A$. But the set $A\cup \text{Lim }A$ is defined as the set $\{x\in X\ |\ x\in A\ \vee\ x\in \text{Lim }A\}$, namely, the set of all $x\in X$ where $x\in A$ or $x\in \text{Lim }A$. Equivalently, th closure of $A$ is the set $A$ and its limit points.

(This is the first example that came to mind, but hopefully you get the idea.)

I know I'm mixing formal logic with natural (human!) language to the point of frivolity, but I'm wondering if there's a non-mundane reason for this equivalence. Perhaps or and and are "dual", in some sense? Or perhaps the equivalence arises just because natural language is really weird?

(One could disregard this question by saying "the meaning of or in logic is not the same as in natural language" (usually natural language uses exclusive-or), just like the meaning of if in logic is not the same as in natural language (usually natural language uses if-and-only-if), but I get the feeling something else is going on.)

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    $\begingroup$ There is duality via deMorgan. Also, e.g., $(p\land q)\to r$ is equivalent to $(p\to r)\lor (q\to r)$ $\endgroup$ – Hagen von Eitzen Jul 15 '17 at 8:36
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    $\begingroup$ That is because the logical 'AND' corresponds to the INTERSECTION in set theory, which in turn corresponds to the 'AND' in the English language. Likewise for the logical 'OR', which corresponds to the UNION in set theory, which corresponds to the 'OR' in the English language. $\endgroup$ – user1952500 Jul 15 '17 at 8:51
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    $\begingroup$ The word and is often used for adding: "3 and 5 is 8", "this things and those things". In mathematics these, especially the last expression, correspond to taking a union, which in logics correspond to or. $\endgroup$ – md2perpe Jul 15 '17 at 12:10
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    $\begingroup$ I think that at least to some extent this question is comparing apples and oranges. (Try to rewrite that in predicate logic!) $\endgroup$ – David K Jul 15 '17 at 16:13
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    $\begingroup$ This is more of an English language question than a math question. Why does the English word "and" sometimes correspond to the logical "or" connective? $\endgroup$ – Tanner Swett Jul 15 '17 at 16:34

10 Answers 10

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Consider: "All fruits and vegetables are nutritious"

Rather than:

$$\forall x ((F(x) \land V(x)) \rightarrow N(x)) \text{ Wrong!}$$

it translates as:

$$\forall x ((F(x) \lor V(x)) \rightarrow N(x))$$

But it also translates as (and is indeed equivalent to):

$$\forall x (F(x)\rightarrow N(x))\land \forall x (V(x) \rightarrow N(x))$$

So now we see that "Fruits and vegetables are nutritious" is really just shorthand for "fruits are nutritious and vegetables are nutritious"

Applied to your case:

"the closure of A is the set A and its limit points."

can be translated as:

$$\forall x ((x \in A \lor x \in \text{Lim} A) \rightarrow x \in Closure(A))$$

or as:

$$\forall x (x \in A \rightarrow x \in Closure(A)) \land \forall x (x \in \text{Lim} A) \rightarrow x \in Closure(A))$$

In other words, the confusion is because of the difference between the disjunction of conditions and conjunction of conditionals.

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    $\begingroup$ I think you should add a paragraph on how this breaks apart when analysing a linguistic “or” and another one on how and why the latter translations really reflect the linguistic usage of “and” instead of the wrong one. $\endgroup$ – k.stm Jul 16 '17 at 17:47
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    $\begingroup$ So I think the answer concerns: $1)$ the way then right-distributes over or (in propositional and predicate logic), so that $(A\vee B) \Longrightarrow C$ equals $(A\Longrightarrow C)\wedge (B\Longrightarrow C)$, and $2)$ how (the universal closures of) these 2 statements translate to natural language ("all A and B are C") $\endgroup$ – étale-cohomology Jul 17 '17 at 6:21
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    $\begingroup$ @étale-cohomology Exactly! Glad I could help :) $\endgroup$ – Bram28 Jul 17 '17 at 12:48
  • $\begingroup$ By they way, the point raised here can be expressed in terms of my answer: “Rather than $F ∩ V ⊆ N$ the sentence All fruits and vegetables are nutritious translates to $F ⊆ N$ and $V ⊆ N$, that is to say $F ∪ V ⊆ N$”. (Yeah, I’m advertising my own answer here – sue me. I put a lot of effort into it and I think I’ve done a good job in the end. Hope you don’t mind, Bram.) $\endgroup$ – k.stm Dec 18 '17 at 22:13
  • $\begingroup$ @k.stm Don't worry, I won't sue you! In fact, i'll give you an upvote; good answer! $\endgroup$ – Bram28 Dec 18 '17 at 22:39
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This is a vague question, of course, but I want to try to give some perspective on it based on some thinking I’ve done lately.

I think this is a matter of implicitly using active / passive, which behave dually in a somewhat precise mathematical manner. First, closely observe how we use active and passive regarding the relation of containment:

Let $X$ be a given set, and let $A$, $B$ and $T$ be subsets of $X$. Then we say

  • $T$ contains $A$ and $B$, when we mean $T \supseteq A ∪ B$, and
  • $T$ is contained by $A$ and $B$, when we mean $T ⊆ A ∩ B$.

This is because we actually mean to avoid repeating some clauses by using the conjunction “and”, in order to shorten sentences. The statements above are (most often) linguistically perceived to be equivalent to respectively

  • $T$ contains $A$ and $T$ contains $B$.
  • $T$ is contained by $A$ and $T$ is contained by $B$.

So, the conjunction “and” implicitly serves as a conjunction of sentences or propositions.

Note: I don’t mean to define mathematical parlance here, I try to observe it.

Now, if we say that “the closure of $A$ is $A$ and its limits points”, I suspect we immediately understand this to say “the closure of $A$ is the set $T$ containing (exactly) $A$ and all limit points of $A$”. By the above observation, that is the set $T$ such that $T \supseteq A ∪ B$ (when $B$ denotes the limit points of $A$) and $T$ contains nothing more (so $T = A ∪ B$).

However, if we use “and” in other contexts as “$T$ is in $A$ and $B$”, we mean “$T$ is contained in $A$ and $B$”, so $T ⊆ A ∩ B$.


Mathematically, this can be interpreted e.g. lattice-theoretically: Let $L = (L, ≤, ∨, ∧)$ be a lattice, $∨$ denoting the join (maximum) of two elements, $∧$ denoting the meet (minimum) of two elements. Then there’s a dual lattice $L^{\mathrm{op}} = (L, ≥, ∧, ∨)$ and $L$ and $L^{\mathrm{op}}$ are antitonely isomorphic as lattices.

In the dual lattice $L^\mathrm{op}$, let’s write for all $x, y ∈ L$

  • “$y ≤'x$” if $y ≥ x$ in $L$,
  • “$x ∨' y$” for $x ∧ y$ in $L$,
  • “$x ∧' y$” for $x ∨ y$ in $L$.

So $L^\mathrm{op}$ has order $≤'$, join $∨'$ and meet $∧'$.

If we introduce parlance for the relation, the antitone isomorphism is reflected in our language. For elements $x, y ∈ L$, let’s say $x$ kills $y$ whenever $x ≤ y$ in $L$. (I like killing here because it's so violently vivid.) As humans, we now also automatically say $y$ is killed by $x$ for $x ≤ y$, that is for $y ≤' x$.

Next, observe that for $a, b, t ∈ L$ $$t ≤ a ∧ b \Longleftrightarrow t ≤ a~\text{and}~t ≤ b.$$ Using our parlance: $$\text{$t$ kills $a ∧ b$} \Longleftrightarrow \text{$t$ kills $a$ and $b$}$$ This justifies $a ∧ b$ to be called “$a$ and $b$”.

And the analogous is true for $L^{\mathrm{op}}$ (where we have to add primes “$'$” to all symbols and replace “kills” by “is killed by”). Therefore, for $a, b, t ∈ L$, we have \begin{align*} \text{$t$ is killed by $a$ and $b$} &⇔ t ≤' a ∧' b' \\ &⇔t ≥ a ∨ b \\ &⇔\text{$t$ is killed by $a ∨ b$}, \end{align*} also justiying $a ∨ b$ to be called “$a$ and $b$”.

The difference lies in which perspective we take on an implicit relation – from above or from below. And it depends on whether we use (implicit) verbs describing the relation in active or passive.

In our case, $L$ is the powerset of some space $X$ with $\supseteq$ as order and killing is containing.

This also explains why such a confusion doesn't happen when lingustically using “or”: We never say “$A$ or $B$” as in “the closure of $A$ is $A$ or its limit points”. This is because $t ≤ a ∨ b ⇔ \text{$t ≤ a$ or $t ≤ b$}$ does not hold for general lattices (think of disjoint union), so there’s no justification for saying “$a$ or $b$” for $a ∨ b$; and there need not be any other element fulfilling this universal property: There is (in general) no set “containing exactly $A$ or its limit points”.

(And in a sense, this generalises to categories with limits and colimits.)

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It depends on how you word it. Does $A \cup B$ correspond to "or" (as the notation $\vee$ for "or" suggests), or does it correspond so "and"?

You argue that $A \cup B$ corresponds to "and", since $$ A \cup B\quad\text{consists of both}\quad A\quad\text{and}\quad B $$

The reasoning I guess used by Boole and his followers is: $A \cup B$ corresponds to "or" since $$ A \cup B = \{x\;:\; x \in A \text{ or } x \in B\} $$

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Common language "do A or B" is interpreted corresponding to formal language "do either A or B". This is symptomatic when scientific patterns are applied to real life phenomena. As you wrote, mixing common and formal language can be confusing.

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You can write $(x \in A) \lor (x \in B) \implies (x \in C)$ or equivalently you can write $(A \subseteq C) \land (B \subseteq C).$

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The same question comes up in computer programming, where new programmers often write incorrect code that sounds correct if you read it aloud and interpret it as an English sentence. The most common mistake is thinking "if x equals 1 or 2, ...", and writing

if x = 1 OR 2

when what they really need to write is

if x = 1 OR x = 2

Different languages have different syntactical variations, but there is a common theme: in the English sentence, the word "or" joins 2 nouns, and the disjunction becomes the object of the verb "equals". The "equals" operator in English can take a numeric object $x$ on one side, and a disjunction of a set of numeric objects $\vee S$ on the other side. It produces a result defined by:

$$\mathrm{equals}(x, \vee S) \longleftrightarrow \bigvee_{i\,\in\,S} \mathrm{equals}(x,i)$$

The OR operators in math and most computer languages don't do anything like that. What does $\vee S$ even mean if $S=\{1,2\}$? What's $1\vee2$? It's a grammatical object that native English speakers can process without any effort, and it sounds like it's made of good, logical, mathematical words, so it's surprising to many people the first time they see a computer fail to interpret it the "obvious" way.

To write the statement correctly using math-style $\vee$ or computer-style OR, you must be able to form an equivalent statement in a restricted English in which "or" can only be used to join an independent clause making a statement of fact with another independent clause making a statement of fact. When you have "and" or "or" joining nouns in your English sentence, you'll be in trouble if you try to blindly translate it word by word into an equation.

In your example "and" is joining "the set $A$" and "its limit points", both of which are noun phrases. That's how you know that's a non-math-y "and".

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In the example, even though the closure of $A$ contains both $A$ and its limit points as sets, a point of the closure does not have to be in both. I think this is the root of confusion in this specific case.

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In English, "AND" has the same meaning as AND. "OR" usually has the same meaning as XOR. If English-speakers wish to emulate the value of the logical OR, they use "AND/OR". The EITHER-OR and NEITHER-NOR constructions in English are obviously XOR constructions. (NEITHER-NOR is equivalent to EITHER X OR Y = NOT-Z.)

As to why this is true, I have no idea.

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    $\begingroup$ Not true. 'Either .. or ...' in English is not necessarily the exclusive or. I can say 'I want to be either rich or happy', and everyone understand that I am just fine with being both rich and happy. ANd the 'neither ...nor' always means neither one, i.e. the negation of the inclusive or: 'I am neither rich nor happy' means 'I am not rich and i am not happy', so that has nothing to do with the XOR. $\endgroup$ – Bram28 Jul 20 '17 at 1:35
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The sense of connectives flips across the boundary between foreground and background (much like the sense of an inequality flips when you multiply it through by -1). For example, if you’re the IT person in your company and the boss says to you, “Give me a list of all our salesreps living in Chicago and Boston.”, then you know that as you pass the salesrep file, you change the boss’s “and” to logical “or”. That is, you output the salesrep’s information if, and only if, the salesrep lives in Chicago OR the salesrep lives in Boston. (The boss was being “illogical”, because the boss is always in foreground, and foreground, the terrain of ‘natural language’, is governed not by logic, but by perception and convenience.) So, a foreground ‘and’ becomes a background ‘or’. Also, a foreground ‘or’ becomes a background ‘and’. For example, if you are the hospitality agent for your company and the boss says to you, “We want to offer each guest the choice of coffee or tea.”, then in background (that is, in the company kitchen) you have to prepare a batch of coffee AND a batch of tea. So, a foreground ‘or’ becomes a background ‘and’. Also, problematical situations are alluded to or acknowledged by using this flip-phenomenon. Specifically, the natural habitat of tautologies is, of course, background. Therefore, foregrounding a tautology (by, for example, saying it in the presence of others) alludes to or acknowledges a problem. For example, it is tautologically true that boys will be boys, but if you happen to overhear someone actually SAYING so, you can be pretty sure it’s not good news that they are commenting on. To take another example, someone wearing a shirt that says, “It is what it is.” is endemically unhappy about something in the world.

(This is a re-post of what I have written here.)

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I think this can be explained in another way which might be a bit easier for one to understand.

Now let's assume we have two independent simple statements:

s1 (e.g. 1+1=2) s2 (e.g. 1+1>2)

And with them, we can generate two compound Statements:

S1: s1 and s2 (e.g. 1+1=2 and 1+1>2) S2: s1 or s2 (e.g. 1+1=2 or 1+1>2)

If S1 is true, both s1 and s2 must be true at the same time, since for the example I gave above, S1 is false. (1+1=2 is true but 1+1>2 is false)

If S2 is true, it means at least one of s1 and s2 is true, which means as long as one of them is true, then "s1 or s2" is true. So for the example above, S2 is true. (1+1=2 is true so regardless of s2’s truthfulness, the Statement “1+1=2 or 1+1>2” is true.) In other words, s1 and s2 must both be false at the same time to output a “false” value of "s1 or s2”.

To put it into more formal and academical words in logics, “And” is referred as “Logical Conjunction” and “Or” as “Logical Disjunction” and part of the Binary Operations, a type of Truth Table.[1]

Hope this helps.

[1]. Anellis, Irving H., Peirce’s truth-functional analysis and the origin of the truth table, Hist. Philos. Log. 33, No. 1, 87-97 (2012). ZBL1270.03008.

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