My logic textbook defines the notion of logic equivalence as:

A proposition $P$ is logically equivalent to a proposition $Q$ (written $P\Leftrightarrow Q$) when the biconditional $P\leftrightarrow Q$ is a tautology.

Using this definition one can easily show that given two propositions $P$ and $Q$ the conditional $P\rightarrow Q$ is equivalent to $\sim P\vee Q$, that is:

$$P\rightarrow Q\Leftrightarrow \sim P\vee Q.$$ But when I try to interpret this with words it seems not to correspond to what my intuition says an equivalence between proposition should mean.

For instance, consider the propositions

$P:\textrm{There is smoke}$

$Q:\textrm{There is fire}$

Then, according to the previous equivalence, the proposition:

$$P\rightarrow Q: \textrm{If there is smoke then there is fire}$$

is equivalent to:

$$\sim P\vee Q: \textrm{There is no smoke or there is fire}.$$

Why does this make sense?


  • $\begingroup$ Does math.stackexchange.com/questions/2723860/… help you? Or one of the other answers to the same question? (The short version of my answer to that question: $P \rightarrow Q$ should be "vacuously true" when $P$ is false, so that when we move into predicate logic, the $x$ with the property $\neg P(x)$ do not interfere with the semantics of $\forall x \: P(x) \rightarrow Q(x)$.) $\endgroup$
    – Ian
    Apr 12, 2019 at 2:15
  • $\begingroup$ It is misleading to say "two propositions are logically equivalent". The relation holds between formulas of propositional logic. Thus, to check equivalence you have to take into account two things: (i) the fact that in prop logic the truth value is the only "meaning" of a propositional letter; and (ii) the def of logical equiv : same truth vale for every truth assignment. $\endgroup$ Apr 12, 2019 at 6:41

3 Answers 3


Part of the problem is that $\rightarrow$ doesn't exactly mean what people think it should mean. For instance $$2 + 2 = 5 \rightarrow 2 + 2 = 4$$ is a true implication, because the consequent $2 + 2 = 4$ is true.

I like to think of $A\rightarrow B$ as if we know $A$ is true, then we may conclude $B$. With this interpretation, there are two options for $A$, either true or false. If $A$ is false, then we don't learn anything about $B$. If $A$ is true, then we also know $B$ because $A\rightarrow B$. Therefore $$ A\rightarrow B \Rightarrow \sim A \lor B$$

On the other hand if $\sim A \lor B$ then we cvan show that $A\rightarrow B$, because if we suppose $A$ is true, then because $\sim A \lor B$, then $B$ is true too. Hence $A$ being true allows us to show $B$ is true. Therefore $$ \sim A \lor B \Rightarrow A\rightarrow B$$


I assume you can write out the truth table and verify the correctness of the equivalence. If you want an intuitive way to understand why it follows, it can be hard to see. The expression $P\rightarrow Q: \textrm{If there is smoke then there is fire}$ means different things all at once.

If there is smoke, there must be fire. This covers that if P is true (there is smoke), then Q must also be true (there is fire), but also that we cannot have the case that P is true (there is smoke) and Q is true (there is no fire)

If there is no smoke, then there may or may not be fire. It is logically true that there being no smoke tells us nothing about where there is fire or not.

Translating this to the second statement, it is saying that we have two options. Either there is no smoke - which in terms of truth statements, it doesn't matter if there is fire or not for it to hold logically true. That is what we said above in the second interpretation.

Or, there is smoke, in which case the statement can only be true if Q is true - there is fire. If there is smoke and there is no fire, the statement is false, which is consistent with saying "We cannot have the case where there is smoke, but no fire".


Classical propositional calculus has only truth-functional connectives. The truth value of a well-formed formula $A@B$ where $A$ and $B$ are well-formed formulas and $@$ is an arbitrary connective is entirely dependenton the truth values of $A$ and $B$ .

We can examine the truth condition of the expressions you have given:

                   P -> Q     ~P \/ Q
        P and Q      T           T
    P but not Q      F           F
    Q but not P      T           T
neither P nor Q      T           T

Since the two propositions have the same truth conditions they are logically equivalent.

Your book is saying that if $A$ and $B$ are logically equivalent then $A \leftrightarrow B$ is a tautology.

$(P \to Q) \leftrightarrow (\lnot P \lor Q)$ is itself a well-formed formula with its own truth conditions.

                   (P -> Q) <-> ~P \/ Q
        P and Q              T
    P but not Q              T
    Q but not P              T
neither P nor Q              T

We can see from the table below that it's a tautology.

It's important to note that "$A$ is logically equivalent to $B$" and "$A \leftrightarrow B$ is a tautology" do not mean exactly the same thing.

The former is a statement about the system we are working in, the latter is a statement in the system we are working in.

The truth functional meaning of $\to$ is somewhat counterintuitive, but that's because if in natural language usually has some sort of causal meaning attached to it.

Propositional calculus is not rich enough to represent anything remotely similar to causality. Everything is either true or false. And a well-formed formula with no free variables is indistinguishable from true or false.

Statements like If I ate 40 donuts yesterday, I would have a stomach ache today. are simply true in this setting, as are statements like If I ate 40 donuts yesterday, then the moon is made of green cheese.

The difference in meaning between the two is not captured by propositional calculus.


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