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In formal/mathematical logic, what are the rules governing "identity" (equals sign) use? Why is it that a conditional (or in certain instances a biconditional) sign is used instead to show a relation between propositions.

I'm a high school student with a developing interest in logic, so any insight into the philosophy guiding the semantics of logical languages would be appreciated.

Thank you for your time.

Edit: Thank you for your responses. I have two fallacious arguments written propositionally: If P, then Q If P, then R Therefore: If Q, then R And If P, then Q If R, then Q Therefore: If P, then R However, if these particular propositions were interpreted as being connected not by a conditional sign but by an "=" (identity) sign, wouldn't we have examples of the transitive property (i.e. P=Q, P=R, so Q=R using the rule of identity elimination in Fitch calculus)? However, neither conditional/biconditional introduction/elimination would be able to prove this in Fitch. Is this why it's a bad idea to use "=" when translating to the more primitive language of propositional logic?

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  • $\begingroup$ Your use of symbols $P$ and $Q$ for propositions suggests an abstract interpretation, in which a proposition is an "atomic" entity (no "internal" structure), in which case it makes sense to consider propositions are either equal (the same) or not equal (different, capable of being assigned independent truth values). $\endgroup$
    – hardmath
    Commented Aug 28, 2016 at 23:49
  • $\begingroup$ Thank you for your responses. I have two fallacious arguments written propositionally: If P, then Q If P, then R Therefore: If Q, then R And If P, then Q If R, then Q Therefore: If P, then R However, if these particular propositions were interpreted as being connected not by a conditional sign but by an "=" (identity) sign, wouldn't we have examples of the transitive property (i.e. P=Q, P=R, so Q=R)? Is this why it's a bad idea to use "=" when translating to the more primitive language of propositional logic? $\endgroup$
    – David Marlowe
    Commented Sep 8, 2016 at 20:09
  • $\begingroup$ No, equality, implication and biimplication are all transitive relations. $\endgroup$
    – Rob Arthan
    Commented Sep 8, 2016 at 20:16
  • $\begingroup$ Or I guess "identity elimination" is what it's sometimes called, the use of the transitive property. $\endgroup$
    – David Marlowe
    Commented Sep 8, 2016 at 20:26
  • $\begingroup$ Thanks, Rob Arthan. I think I do understand why they're transitive--but the using the conditional in that case certainly creates a fallacy $\endgroup$
    – David Marlowe
    Commented Sep 8, 2016 at 20:30

5 Answers 5

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There is nothing ill-conceived about thinking of what is usually written as $P \Leftrightarrow Q$ or $P \leftrightarrow Q$ as an equation between the propositions denoted by $P$ and $Q$. However, it is conventional in mathematical logic not to use the equals sign for this relation.

This convention avoids various kinds of ambiguity and fits in well with the concepts of first-order logic (where we have a specific universe of discourse in mind, like natural number arithmetic, in which equality denotes equality of natural numbers, while bi-implication denotes equivalence of propositions about the natural numbers).

There are higher-order logics in which the world of propositions forms part of the universe of discourse and then bi-implication just becomes a special case of equality.

There is a separate question of notation for syntactic identity between linguistic constructs. I.e., the notation to use when we are thinking of $1 + 2$ or $A\land B$ as sequences of symbols not expressions that denote numbers or truth values. The symbol $=$ is often used for syntactic identity, but this usage can give rise to confusion: using $\equiv$ or $:=$ is a common way of avoiding this.

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  • $\begingroup$ So CpCqp = CCpCqrCCpqCpr, CCpqCCqrCpr=CCNppp, ApNp=CCpCqrCqCpr and so on for all pairs of tautologies? $\endgroup$
    – Doug Spoonwood
    Commented Aug 29, 2016 at 0:59
  • $\begingroup$ It's also convenient to have $\iff$ as a more primitive notion than $=$, we can define $=$ to be the set of pairs $(a,b)$ such that $P(a)\iff P(b)$ for all predicates with a parameter. This simplifies the logical language a bit, I believe. $\endgroup$
    – YoTengoUnLCD
    Commented Aug 29, 2016 at 1:37
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    $\begingroup$ @YoTengoUnLCD: yes that's Leibnitz's law of the identity of indiscernibles. I don't think it makes any simplifications to the language of higher-order logic. $\endgroup$
    – Rob Arthan
    Commented Aug 29, 2016 at 1:44
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    $\begingroup$ @DougSpoonwood: yes, when translated out of the ancient notation that you insist on using, classical logical equivalences are equations (identities) over boolean algebras. $\endgroup$
    – Rob Arthan
    Commented Aug 29, 2016 at 2:01
  • $\begingroup$ @RobArthan Isn't that actually grouping all tautologies under an equivalence class, and then just checking equality between the classes of two distinct tautologies? $\endgroup$
    – YoTengoUnLCD
    Commented Aug 29, 2016 at 22:41
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The equality sign could be used between propositions, but they would need to be equal, that is, they would need to be the same proposition.

One reason for using biconditional is that it can link propositions that are not equal, but which have the same truth value. For example, "English has 26 letters" and "Hydrogen has one electron" are both true, but they are not the same. So we would not write an equal sign between them, but we could write a biconditional between them.

In propositional logic, we are often much more concerned with whether propositions have the truth value than whether they are the same - to the point that the equality sign is often left out altogether.

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    $\begingroup$ +1. For the OP, here's an example in pure propositional logic: consider the sentences "$A$" and "$A\vee A$." They are definitely equivalent, but are the "the same" sentence? I would say no - they are different strings of symbols. $\endgroup$
    – Noah Schweber
    Commented Aug 28, 2016 at 23:43
  • $\begingroup$ "English has 26 letters" I suppose that holds if lower case letters and upper case letters in modern (or it looks like Olde also) English have their usual correspondence. But, a and A, at the very least, are not merely size transformations of each other. Nor are q and Q. Nor are r and R. though maybe the above sentences weren't in english. MAYBE ONLY THESE LAST TWO SENTENCES ARE IN ENGLISH. $\endgroup$
    – Doug Spoonwood
    Commented Aug 28, 2016 at 23:54
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    $\begingroup$ So, apparently, both Carl and Noah are unhappy that $1 + 2 = 2 + 1$ and that $0 = 0 + 0$. Hmm? $\endgroup$
    – Rob Arthan
    Commented Aug 29, 2016 at 0:04
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    $\begingroup$ @RobArthan Good point, I was being sloppy. For me the difference is this: we largely agree that the purpose of a string like "$0+1$" is to denote a number, and (with minor exceptions) there's rarely any serious confusion over what sort of number such strings can denote. By contrast, I'm not at all convinced that the string of symbols "$A\vee A$" denotes merely its truth value (I'm also not convinced that this isn't the case), and indeed the question of denotation here is quite messy. So for me that's the distinction - when the kind of denotation is unclear, I go strict. $\endgroup$
    – Noah Schweber
    Commented Aug 29, 2016 at 3:02
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    $\begingroup$ @CarlMummert What does it mean "to evaluate to the same value"? $\endgroup$
    – YoTengoUnLCD
    Commented Aug 29, 2016 at 3:21
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If you mean you find it more sensible to use equality instead of $\leftrightarrow $ with the same purpose, it is mostly a matter of convenience; we already use the symbol of equality for other purposes.

If you mean you want the proposition "P = Q" to mean something like "P and Q are identical": in typical logical languages, formulas or propositions range over whatever the domain of the model is. For example, if we are considering the first order logic of graph theory (i.e. with one binary relation), a formula should have a truth value based on what graph it is interpreted in, and what vertices are assigned to its free variables.

If we allow a construction like "if P, Q are formulas, then P = Q is also a formula", then the interpretation of this sentence would have to depend not on the structure of a graph, but on the syntactic information of the formulas P and Q. This is simply not what we want the propositions to be "allowed" to express.

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    $\begingroup$ +1: It looks to me that it was Carl rather than the OP who raised the spectre of syntactic identity. $\endgroup$
    – Rob Arthan
    Commented Aug 29, 2016 at 0:46
  • $\begingroup$ @Rob The phrase "what are the rules governing 'identity'" made me think that perhaps OP was asking why the equals sign was allowed between terms, but not between formulas. $\endgroup$
    – Mees de Vries
    Commented Aug 29, 2016 at 0:50
  • $\begingroup$ I was supporting your answer, not criticising it. $\endgroup$
    – Rob Arthan
    Commented Aug 29, 2016 at 1:41
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First-order logic already uses = for identity on the domain under discussion (sets, integers or whatever). So even though we could conceivably use = for identity on truth values, it's clearer to have a different symbol $\leftrightarrow$.

Plus the notation $P \leftrightarrow Q$ has the advantage of suggesting $P \rightarrow Q$ and $Q \rightarrow P$.

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It looks like the first person to use the '=' sign was Robert Recorde in 1557. In translation Recorde says "And to avoid the tedious repetition of these words : is equal to : I will set a pair of parallel lines thus: =, because no two things can be more equal."

So one might ask, if you wrote an equals sign between logically different, yet truth-functionally equivalent, propositions would they qualify as two things which can't be more equal? Having studied the consequences of the forms of certain propositions, I've found that the consequences of those forms can end up as very different even when those forms qualify as having the same truth-value. So, I do not think it correct to write '=' between propositions.

A distinct answer: P and Q are not propositions, unless specified and oftentimes they are not specified. They serve as variables for propositions. Writing p = q or equivalently a = b or equivalently c = f would literally mean that the variables were equal.

There is no case where P and Q qualify as propositions. It just doesn't qualify as coherent or makes for wording that could use revision for clarity. P and Q may represent propositions, but that is all. Considering P and Q as propositions makes for a counterfactual situation.

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    $\begingroup$ Not sure why the downvote -- it's a good answer with informative historical context. $\endgroup$
    – Caleb Stanford
    Commented Aug 30, 2016 at 18:22
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    $\begingroup$ I didn't downvote, but I disagree with the answer. The = symbol, were it to be used here, would assert that the two propositions have the same truth value (or truth values that couldn't be any more equal, if you prefer) in the model under consideration, not that they have the same consequences from a proof-theoretic view. $\endgroup$
    – user8651
    Commented Aug 30, 2016 at 18:31

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