We can define using the '$\equiv$' an 'identity', for example the trigonometric one I gave in the title. Given two expressions connected by triple bar, for example, $\sin(A+B) \equiv \sin(A)\cos(B)+\cos(A)\sin(B),$ can we define the identity as being equivalent to the quantified statement: $$\forall A\forall B(\sin(A+B) = \sin(A)\cos(B)+\cos(A)\sin(B))?$$

The difficulty I have is say we have two angles and the relation is such that $A$ is a function of $B$ depend on each other, we should in theory be able to add them no matter what their values may be at any time, so the use of the ≡ symbol to signify that the two sides will be equal no matter what values they will have at any time (for each $A$ there is now a corresponding B as $A=f(B)$ for some function $f$). However, the double universal quantifier doesn't seem like a statement we can make as for any $A$, it is not possible for $B$ to vary in its entire range in the domain of discourse, as $A$ is a function of $B$ there can only be perhaps one or two possible values of $B$, is the quantified statement simply incorrect or is there a way to get past this?

As an addition, perhaps it is fine to use universal quantification because it is about truth if the variables were able to take those values, ignoring if we have certain restrictions that may not allow a certain combination of values to take place.

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    $\begingroup$ If $\forall A\forall B(P(A,B))$ is true then so is $\forall A(P(A,f(A))).$ I don't see the difficulty. If you think you can go back to the original trig identity and say that it only applies to $A$ and $B$ subject to certain constraints, you're no longer talking about the original trig identity but rather you have some weaker statement. $\endgroup$
    – David K
    Jun 30 at 21:39
  • $\begingroup$ hi user37577. indeed, the way you have quantified it is absolutely correct! and it is the case that "for any $A$ it is possible for $B$ to vary in its entire range in the domain of discourse". why do you have the impression that "$A$ is a function of $B$"? what function $f$ do you have in mind? $A$ and $B$ in this statement are completely independent $\endgroup$ Jun 30 at 21:40
  • $\begingroup$ Universal quantifiers commute, so you don’t have to worry about order in choosing $A$ and $B$. You could write $\forall A,B$ or $\forall B\forall A$ and have the same meaning. $\endgroup$ Jul 1 at 12:10
  • $\begingroup$ Identities are universally quantified: $\forall x \forall y [(x+y)^2=x^2+y^2+2xy]$ $\endgroup$ Jul 1 at 15:04
  • $\begingroup$ @AtticusStonestrom It could be any $f$, $A=2B$ $A=B^2$ any of your choosing, this would mean that B can only have values in its range for certain A. $\endgroup$
    – user1047679
    Jul 4 at 8:59

1 Answer 1


we can define using the '$\equiv$' an 'identity', for example the trigonometric one I gave in the title

Trigonometric identities are not identities-by-definition.

Also worth noting: the term ‘identity’ has contrasting usages in mathematics; as does the $\text‘≡\text’$ symbol.

$A$ is a function of $B$ depend on each other,

There is no dependency between $A$ and $B$ (as explained in our previous discussion), because the title equation’s solution set is the entirety of $\mathbb R^2,$ not a proper subset of $\mathbb R^2.$ It is precisely because that equation is universally quantified that we, by definition, call it an identity. (This pertains to the second bullet point of our other previous discussion.)

On the other hand (the previous link's first bullet point), if a two-variable equation (e.g., $xy=10)$ is satisfied by some but not all points in $\mathbb R^2,$ then its solution set’s $x$- and $y$-coordinates generally do depend on each other, and it is a conditional equation instead of an identity.


  • That equation $P$’s solution set’s $x,y$ are independent typically corresponds to $$\forall (x,y) \; P(x,y)$$
  • That equation $P$’s solution set $S$’s $x,y$ are dependent typically corresponds to $$\forall(x,y){\in}S \; P(x,y)$$
  • $\begingroup$ In my case they would be satisfied by all values of the variables, however if they cannot actually in practice ever be able to have that combination of values without invalidating another relation that I wish to be true if they are to be functions of one another. $\endgroup$
    – user1047679
    Jul 4 at 9:02
  • $\begingroup$ 1. There's no need for "would": your specified identity is satisfied by all $(A,B)$ couples. $\quad$ 2. The rest of your sentence is grammatically incoherent; in any case, your specified identity invalidates no correct relation, and $A$ and $B$ aren't functions of each other, and that's that about that. $\quad$ 3. You want to additionally impose a constraint $A=f(B)$ on that equation? Then it's no longer an identity, and my Answer's final line explicitly explains how you might formalise it as a conditional equation. $\endgroup$
    – ryang
    Jul 4 at 10:30
  • $\begingroup$ Conditional equation, so the use of triple bar is not correct when we impose a constraint, I thank you. $\endgroup$
    – user1047679
    Jul 4 at 15:50

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