3:08 PM
Suppose you have some premises that may be replaced with some other premises--not necessarily the same number of them--such that all the same conclusions can be inferred from each collection of premises. That is, suppose that the sentences $S_1, S_2, \ldots, S_n$, taken together, imply all the same conclusions (which just means, imply all the same sentences) as are implied by the sentences $T_1, T_2, \ldots, T_m$, taken together.
Another way to describe this situation is that when the sentences $S_1, S_2, \ldots, S_n$ appear as (some or all of the) premises of an argument, the validity of the argument is always unaffected by (i.e., invariant under) the operation of replacing them with $T_1, T_2, \ldots, T_m$. No matter what other premises the argument has and no matter its conclusion, if it was valid before, it still is, and if it was invalid before, it still is.
Then we say that $S_1, S_2, \ldots, S_n$ and $T_1, T_2, \ldots, T_m$ are equivalent collections of premises. Notice that this happens iff each collection of premises implies all the individual premises of the other collection, and thus implies their conjunction, i.e., when it is both the case that $S_1 \wedge S_2 \wedge \ldots \wedge S_n$ implies $T_1 \wedge T_2 \wedge \ldots \wedge T_m$ and that $T_1 \wedge T_2 \wedge \ldots \wedge T_m$ implies $S_1 \wedge S_2 \wedge \ldots \wedge S_n$.
That, in turn, happens when, and only when $$(S_1 \wedge S_2 \wedge \ldots \wedge S_n) \leftrightarrow (T_1 \wedge T_2 \wedge \ldots \wedge T_m)$$ is valid. (If it is not valid, then there is some interpretation under which $S_1 \wedge S_2 \wedge \ldots \wedge S_n$ and $T_1 \wedge T_2 \wedge \ldots \wedge T_m$ have different truth values.)
In the simple case of $n = m = 1$ we have the equivalence of the single premise $S$ to the single premise $T$. (Note that we can obtain this by replacing the separate sentences $S_1, S_2, \ldots, S_n$ with $S_1 \wedge S_2 \wedge \ldots \wedge S_n$ and replacing $T_1, T_2, \ldots, T_m$ with $T_1 \wedge T_2 \wedge \ldots \wedge T_m$.)
Thus equivalence of a single sentence $S$ to another single sentence $T$ is validity of the biconditional $S \leftrightarrow T$. Just as implication between two individual sentences is validity of the conditional, equivalence of two individual sentences is validity of the biconditional.
Analogously to the relationship between implication and the conditional, equivalence isn't the biconditional, because $S \leftrightarrow T$ is true whenever $S$ and $T$ happen to have the same truth value, but for $S$ to be equivalent to $T$ means that $S \leftrightarrow T$ is not merely true but valid--that is, true under all interpretations, true by virtue of the logical makeup of $S$ and $T$.
Besides being analogous in that way, there is a very important relationship between implication and equivalence. $S$ is equivalent to $T$ iff both $S$ implies $T$ and $T$ implies $S$.
As with implication and the conditional, some people use "equivalence" to mean the biconditional, and say "is equivalent to" in place of "if and only if." This usage has the same motivations, carries the same disadvantages, and is similarly popular among mathematicians--and I don't recommend it, either. (Though it is hard to avoid in some popular constructions used in mathematics writing, such as "The following are equivalent...".)
Just as the ambiguity about implication can be resolved by calling validity of the conditional logical implication and calling mere truth of the conditional material implication, the corresponding ambiguity about equivalence can be resolved by calling validity of the biconditional logical equivalence and calling mere truth of the biconditional material equivalence. Two sentences with the same truth value are materially equivalent, but they are very often still not logically equivalent.
When I talk about implication (of sentences or schemata) and I do not qualify it, I mean logical implication. Likewise, when I talk about equivalence (of sentences or schemata), I mean logical equivalence.
Notice that you can freely interchange a sentence or schema with an equivalent one, even when the sentence or schema you start with is part of a larger sentence or schema. This preserves interesting properties such as truth, validity, and what can be inferred.
That is, suppose $S$ is a sentence or schema with $T$ appearing as any distinct part of it, and $T$ is equivalent to $T'$. Denote the sentence or schema obtained from $S$, by replacing $T$ with $T'$, as $S'$. Then $S$ is equivalent to $S'$. For example, as you found, $p \rightarrow q$ is equivalent to $\neg p \vee q$. So $(p \rightarrow q) \wedge r$ is equivalent to $(\neg p \vee q) \wedge r$.
Especially when applying this to ordinary language, some care has to be taken in deciding what qualifies as what I have called a "distinct part." For example, "Smith is away" is equivalent to "Smith is away and Smith is away," but "'Smith is away' is 11 letters long" is not equivalent to "'Smith is away and Smith is away' is 11 letters long".
I mean that they are not logically equivalent. But actually they don't even happen to have the same truth value (i.e., the biconditional does not hold between them, i.e., they are not even materially equivalent).
In that example, the problem is easy to spot because of the explicit quotation. "'Smith is away' is 11 letters long" does not, logically speaking, contain "Smith is away" as a clause. When one quotes text, one forms a term that names that text. "'Smith is away'" is a name for "Smith is away".
But most people, including me in these messages, don't always show quotation when it is happening. Consider, for example, the sentence, "$S$ implies $T$ if and only if $S \rightarrow T$ is valid." This is correct, as I intend it and as I think you understand it. But in the first clause, "$S$ implies $T$", I am using "$S$" and "$T$" as names. In contrast, in the second clause, I am doing something more subtle: I intend that the reader understand their presence to signify substitution.
In that particular context, I intend that "$S \rightarrow T$" be construed as a simpler way to say "the result of concatenating $S$ (i.e., that which '$S$' refers to, which is presumably not the single symbol '$S$'), the symbol '$\rightarrow$', and $T$ (i.e., that which '$T$' refers to, which is presumably not the single symbol '$T$'), in that order." This is known as quasi-quotation.
Note that, in typical use, formulas and sentences built up with sentential connectives like "$\rightarrow$" do not involve quasi-quotation. "$\neg$", "$\wedge$", "$\vee$", "$\rightarrow$", and "$\leftrightarrow$" are sentential connectives, not predicates; syntactically, they connect actual sentences, not names of sentences.
I mentioned that $S$ is equivalent to $T$ iff both $S$ implies $T$ and $T$ implies $S$. This relates to how "$\wedge$" interacts with validity. $S_1$ and $S_2$ are both valid iff $S_1 \wedge S_2$ is valid. You can think of this as validity distributing over conjunction.
(I say it that way, rather than saying "validity distributes over conjunction," because there is some implicit quotation and concatenation going on behind the scenes.)
Note that this works with conjunction, but it does not work most connectives. For example, it does not work with disjunction: $p \vee \neg p$ is valid, but it does not follow from this that $p$ is valid or $\neg p$ is valid, and neither is.
So suppose $S$ implies $T$ and $T$ implies $S$. Then by the definition of implication, $S \rightarrow T$ is valid (because $S$ implies $T$) and $T \rightarrow S$ is valid (because $T$ implies $S$). Therefore their conjunction, $(S \rightarrow T) \wedge (T \rightarrow S)$, is valid.
But that is equivalent to $S \leftrightarrow T$. You can just tell that this is so, and you may even have defined "$\leftrightarrow$" this way. But in the interest of rigor, note that this is a truth-functional validity and can be verified by a truth table.
Since $(S \rightarrow T) \wedge (T \rightarrow S)$ is valid and equivalent to $S \leftrightarrow T$, and validity is invariant under interchange of equivalents, $S \leftrightarrow T$ is valid. Then by the definition of equivalence, $S$ is equivalent to $T$. These steps are reversible, so going the other way--inferring from "$S$ is equivalent to $T$" that "$S$ implies $T$ and $T$ implies $S$"--is similar. Therefore $S$ is equivalent to $T$ iff both $S$ implies $T$ and $T$ implies $S$.
As an aside, one thing that proof pragmatically illustrates is why, when talking about validity, implication, and equivalence, I've chosen to use letters like $S$ and $T$ to signify or hold the place of sentences or schemata that may have arbitrarily complex internal structure that matters for what I'm talking about... rather than using sentence letters like $p$ and $q$.
After all, truth-functional schemata like $p \leftrightarrow q$, which comes out false when $p$ is true and $q$ is false as well as when $p$ is false and $q$ is true, are not valid. (But don't get me wrong: $(p \rightarrow q) \wedge (q \rightarrow p)$ is equivalent to $p \leftrightarrow q$, so $[(p \rightarrow q) \wedge (q \rightarrow p)] \leftrightarrow (p \leftrightarrow q)$ is valid.) If I'd explicitly quasi-quoted things I wrote as compounds of $S$ and $T$, this may have been less confusing.
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