How to Read Mathematical Formulas In order to understand the mechanics of mathematical formulas, it's important toknow the conventions regarding the order in which logical operators apply. As aquick summary, the precedence order for applying logical connectives is as follows: ∀𝐿 ( ) …. ( ) (∃𝐿)(….) Quantifiers are the words that bind tightly, that hold things together. A quantifierapplies to whatever comes immediately after it. Typically, what comes after the word"then" is a group of other items, such as "and's and or's and not's." You would putthose items in parentheses, brackets, square brackets or whatever. Red is apredicate that applies to all balls. If we said, "For all balls Red B," then "Red" wouldbe the subject and "B" would be the predicate. ∀𝐵 𝑅𝑒𝑑 (𝐵) As shown by the following example, the for all relation applies only to sets that aredefined as a subset of their containing set. If we wanted the for all relation to apply tothat set, we would need to include parentheses around it. (∀𝐵) 𝑅𝑒𝑑 (𝐵) The quantifier "not" is used to negate a statement. When negating a statement, thenegation symbol followed by parentheses should be placed immediately after thequantifier, and everything between the parentheses should be negated. For example,if you want to say that 3 is neither greater than 0 nor less than 0, then you wouldwrite "not (3>0) and not (3<0). ¬(3 > 0 ∧ 3 < 0) If 3 is greater than 0, and 3 is less than 0, then the conjunction of these twostatements is false. But if the conjunction is false, then its negation must be true. Sothis sentence is true. But suppose we had written it differently: The case in which 3 isgreater than 0 and 3 is less than 0. In this case, both clauses are true but the entirestatement is false because they are mutually exclusive and jointly exhaustive—onecannot be true without the other one being false. ¬(3 > 0) ∧ (3 < 0) T If the parentheses did not include both of those, then what we have is aconjunction—a set of statements that are true together. In this case, the twostatements are both false, so we have something that is false. ¬((3 > 0) ∧ (3 < 0)) F F So these are clearly not the same, because this one is true and that one is false.Here, the negation applies to everything between them, making it true. ¬(3 > 0) ∧ (3 < 0)
The issue between these two sentences wasn't whether there were parenthesesaround the 3 greater than or less than 0. The issue was whether the parenthesesgoverned everything that was next, or just the one thing that was next. These are notthe same. ¬((3 > 0) ∧ (3 < 0)) ≢ ¬ 3 > 0 ( ) ∧ 3 < 0 ( ) The next one is a conjunction: 3 is greater than 0, and 3 is less than 0. 3 > 0 ∧ 3 < 0 The fact that 3 is greater than 0 and that 3 is less than 0 can be expressed using theconjunctions "greater than" and "less than," which are known as quantifiers. (3 > 0) ∧ (3 < 0) Some mathematicians argue that conjunction and disjunction have the samestrength. Some people say that conjunction is stronger than disjunction, but thisargument is not particularly strong. In any case, it is best practice to always useparentheses when you want to disambiguate yourself from your statement. The pointis that whenever one should apply a grouping rule such as conjunction or disjunction,one should always use parentheses around it. (…..) ∨ (….) And likewise, if you have an implication or a conditional, the whole thing will be theantecedent and this will be the consequent. (…) ⇒ (…) Now, in here, there may be all sorts of conjunctions and disjunctions and stuff. Inhere you may find quantifiers, negation signs and a whole lot more. So wheneveryou look into the sort of basic thing with all of these is when you've got a quantifier ora negation symbol or a conjunction or a disjunction or an implication. ⇔ If one assumed that A/C had a tighter binding than B/D, they could write somethinglike A and B or C and D. 𝐴 ∧ 𝐵 ∨ 𝐶 ∧ 𝐷 And if you put space in there, it is fairly clear that it meant to be either A or B, or C orD. (𝐴 ∧ 𝐵) ∨ (𝐶 ∧ 𝐷) The golden rule is to put parentheses around the quantifiers. (∀𝐿) ∃𝑆 1 ( ) 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆) ⇒ ∀𝑆 2 ( ) 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆 2 ) [ ] Now, the parenthetical phrase there teams up with the parenthetical phrase there.And we've them not as parentheses but as square brackets to make it absolutelyclear.
(∀𝐿) ∃𝑆 1 ( ) 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆 1 )⇒ ∀𝑆 2 ( ) 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆 2 ) [ ] The argument is going to say that, for any given licence L, if there exists at least onestate in which the licence L is valid, then it will hold that L is valid in every possiblestate. For every licence, L, if a valid state is any state in which L is valid, then thereexists a state where L is valid. This second sentence simply applies what waspreviously said to this particular situation. So it literally says: "For every licence, if thevalid state is a state in which that licence is valid, then the exists here simply appliesto this." The exists binds what's next to it: "There exists a state where…" And whatthat means is that there must be a state where the given licence is valid. (∀𝐿) ∃𝑆 1 ( ) 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆 1 )∧ ∀𝑆 2 ( ) 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆 2 ) [ ] What is the difference between the two? Well, in this sentence the for all applies toeverything inside the parentheses. This sentence has a conjunction, so let's seewhat that would mean. In this sentence, the binding is the same: the for all applies toeverything in parentheses. This exists applies to this thing; and this for all applies tothis thing. In both cases, L is still bound—which means that as was determined in (1)above, L inside here is bound. For every licence L, there is a state in which thelicence is valid, and there is a state in which the licence is valid in all states. For anylicence L, there is a state in which L is valid and there is a state in which L is valid inevery state. But this does not mean that the same applies to every single licence. It is important to know that not every license is valid. For example, if you go toCalifornia and try to drive with too much alcohol in your bloodstream, you will findyourself with an invalid license. It is the first thing that was a problem. For everylicence, there is a state in which it is valid. Well, that's simply not the case. Alreadythe first conjunct makes it invalid. The part that says it is valid in a state was theantecedent; therefore, if it is valid in a state, then it is valid in all states. So that said,for every licence—if it is valid in a state—this says that all licences are valid in somestates. But this is not always true; not all licences are valid. Therefore, there is adifference between these two and the difference is meaningful in terms of validity oflicenses and so forth. (∀𝐿) ∃𝑆 1 ( ) 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆 1 ) ⇒ ∀𝑆 2 ( ) 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆 2 ) We don't have these brackets here, so let's look at what's going on. For everylicence there is a state that it is valid in. There is a line here. The for all and theexists do not apply to that line; they apply to what comes next. And there was nobracket, so it does not include here. So what this actually says is that for everylicence there is a state in which the licence is valid. So what this really says is that alllicences are valid somewhere. That isn't true, and it isn't what we want to say either;if that were the case then for all states 2 this would say that L is valid in all states 2. In fact, as we'll see later, there are many kinds of licences that aren't valid in anyjurisdiction. So the antecedent of this conditional is false and the conditional itself isundefined. If you know what L is, you can assign meaning to it. And once you'veassigned a meaning to it, if that L refers to my license and if that L is my license then
we would have a true conditional. But as it stands, you've just got an unbound or freevariable—it doesn't have any real meaning until you assign a value to it. (∀𝐿)(∀𝑆 1 )(∀𝑆 2 ) 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆 1 ) ∧ 𝑉𝑎𝑙𝑖𝑑(𝐿, 𝑆 2 ) [ ] Thus, for every licence and for all pairs of states, the license is valid in one state andthe licence is valid in two states. That means all licences are valid in all states.Again, this is not the case in the United States because you can have invalidlicenses so we've got something that's actually false. There's redundancy herebecause the second S adds nothing new. It simply says that for all licences and forall states, the licence is valid in that state and it's valid in the other state. So we couldjust scrap that and scrap that bit and we'd have the meaning without any of that stuff. In order to distinguish between cases that are commonly confused by beginners, letme write down a transcription of what it means. In short, for all x, if P(x) then Q(x). IfP is true then Q is true. (∀𝑥) 𝑃(𝑥) ⇒ 𝑄(𝑥) [ ] For every real number, if that number is nonnegative, then it has a square root.Thisoccurs frequently in mathematics.Once you've established that "for all" there is arelationship between P and Q. (∀𝑋) 𝑃(𝑥) ∧ 𝑄(𝑥) [ ] It says that every x satisfies P and Q. That's a fairly strong statement, because it onlymeans that if something is an x, then it also satisfies P and Q. So it's not toofrequent, because you could equally well say for all x P(x) and for all x Q(x), which isequivalent to saying everything satisfies P and everything satisfies Q. ∀𝑥𝑃(𝑥) ∧ ∀𝑥𝑄(𝑥) Notice that the statement above is valid, because of the binding from the quantifierfor all to what's next to it. The quantifier for all cannot bind anything else in this case.This is because in mathematics, such a statement would never appear unless it wereconfined by parentheses indicating which variables are being quantified over;however, since an overlap would be confusing here and therefore impractical, thisstatement does not require any additional parentheses. (∃𝑥) 𝑃(𝑥) ∧ 𝑄(𝑥) [ ] This says that for any x, if P of x and Q of x, then there is an x that satisfies both Pand Q. This is quite a strong statement, as it says you can find a single x that hasboth property P and property Q. So you can find an x that has the property P andhas the property Q. So it's strong simply because it makes a claim about all xsatisfying both properties. (∃𝑥) 𝑃 𝑥 ( ) ⇒ 𝑄(𝑥) [ ] There is at least one x such that if P of x, then Q of x. If you see yourself writing anexistence with an implication, the chances are very high that you've sort of gotconfused. This says for every x if it satisfies P, then it satisfies Q. Now that's making
a strong statement. For every x, there's an implication. This simply says, there's onex for which there's an implication. Well, in a sense, it's almost vacuous then. Onething to say for example is that if you can find an x that does not satisfy P, in otherwords, you can find an x anywhere for which P of x is false. If you can find an xanywhere for which P of x is false then you have a conditional that's necessarily true. If you're trying to make an existence statement, then you can do so simply by findingan x that does not satisfy P. Because if you can make that part false, the conditionalbecomes true. So if you're trying to make a strong statement, forget that one. On the other hand, if you see an implication following the word exists and anexistence claim comes before it, then chances are you've gotten confused. Ingeneral, these two symbols are significant: If a formula has either of these twosymbols, it is usually significant because it reduces to two separate claims orpossibilities. However, this one is less significant because it really just reduces to thetwo separate things. This one is also pretty weak because it really just restateswhat's already stated in the previous sentence. Therefore, if you see an implicationfollowing exists with an existence claim before it in your own writing, flag these partsand ask yourself if what you meant is actually there.