Are Implication and Equivalence Interchangeable in Logic?

[Heisenberg7]

I've seen a lot of people use implication and equivalence logic incorrectly. For example, when solving equations (i.e. $x - 2 = 3 \implies x = 5$). Implication is not reversible, thus it only works in one way. By saying, $x - 2 = 3 \implies x = 5$, you are essentially saying that it is unknown whether or not $x = 5 \implies x - 2 = 3$, which is incorrect. In this case, one should use equivalence because that tells us that the first equation gives us the second one and vice versa. To conclude, the correct way would be $x - 2 = 3 \iff x = 5$.

Implication is not reversible and equivalence is. Or perhaps some people use it to denote the next step when solving equations, but in either case, it would be better to use equivalence. I'm curious what other's take might be on this topic.

 

[FactChecker]

I've seen a lot of people use implication and equivalence logic incorrectly. For example, when solving equations (i.e. $x - 2 = 3 \implies x = 5$). Implication is not reversible, thus it only works in one way. By saying, $x - 2 = 3 \implies x = 5$, you are essentially saying that it is unknown whether or not $x = 5 \implies x - 2 = 3$,

I disagree. It says nothing about the reverse implication, no matter what is actually known about it. The only thing the use of the one-way implication says is that he is using that implication. It says nothing more. Anything else is your own supposition -- right or wrong.
We are free to use one logical fact without stating all known logical facts.

 

[Heisenberg7]

I disagree. It says nothing about the reverse implication, no matter what is actually known about it. The only thing the use of the one-way implication says is that he is using that implication. It says nothing more. Anything else is your own supposition -- right or wrong.
We are free to use one logical fact without stating all known logical facts.

Well, the problem is; addition, subtraction, multiplication and division are all well defined operations when it comes to equations, and knowing that, we do know that implication works both ways in this case. By using equivalence, we'd get a correct statement leaving no room for questioning our reason. Meanwhile, implication is only one way. By saying, A implies B, it doesn't mean that B implies A. I am all for being precise in our reasoning (well defined).

 

[Frabjous]

Well, the problem is; addition, subtraction, multiplication and division are all well defined operations when it comes to equations

Division by zero is not well defined.

Most of the time we are trying to get from A to B, so nobody cares about B to A.

 

[Heisenberg7]

Division by zero is not well defined.

Most of the time we are trying to get from A to B, so nobody cares about B to A.

Well, yes and I know that this is probably not a necessary discussion and you could probably go through your whole life without worrying about which one to use, when writing proofs I would say that it is important to know the difference. Especially when your professor is, for example, picky about generality. Either one works, but implication is definitely weaker. (personal experience)

I probably made a mistake by saying that it's incorrect to use implication, but it is weaker.

 

[FactChecker]

Well, yes and I know that this is probably not a necessary discussion and you could probably go through your whole life without worrying about which one to use, when writing proofs I would say that it is important to know the difference. Especially when your professor is, for example, picky about generality. Either one works, but implication is definitely weaker. (personal experience)

I probably made a mistake by saying that it's incorrect to use implication, but it is weaker.

Using the one-way implication allows us to state what is known without worrying about proving or disproving the converse. In general, the converse is not as obvious as it is in your example. It also makes it clear that only the one-way implication is involved in the current proof where including the converse can be misguiding.

 

[fresh_42]

I've seen a lot of people use implication and equivalence logic incorrectly.

Where? I have seen thousands of pages in textbooks and never seen either of the signs. They simply do not occur.

For example, when solving equations (i.e. $x - 2 = 3 \implies x = 5$). Implication is not reversible, thus it only works in one way.

It actually is reversible: $x-2=3 \Longleftrightarrow x=5.$

By saying, $x - 2 = 3 \implies x = 5$, you are essentially saying that it is unknown ...

or far more often, that I do not care as it is irrelevant to my conclusions

... whether or not $x = 5 \implies x - 2 = 3$, which is incorrect.

No. It is not incorrect. $( x=5 \Longleftrightarrow x-2=3) \Longrightarrow (x=5 \Longrightarrow x-2=3 y).$
Noting only the one relevant of two possible cases is a service for the reader. It means that he is not pestered with what is not necessary.

In this case, one should use equivalence because that tells us that the first equation gives us the second one and vice versa. To conclude, the correct way would be $x - 2 = 3 \iff x = 5$.

There is no case.

If in the case of a homework problem where students start with what has to be shown and end up at $1=1,$ however, it is indeed necessary to observe equivalence in every single step. If in case you want to write the solution correctly, you start with $1=1$ and end up with what has to be shown using only implications, regardless of whether equivalences hold. And in the case of a book or a paper, you do not even use symbols like $\Longrightarrow$ or $\Longleftrightarrow$ and use language instead.

Implication is not reversible and equivalence is. Or perhaps some people use it to denote the next step when solving equations, but in either case, it would be better to use equivalence. I'm curious what other's take might be on this topic.

A final note: even logic does not use the symbols $\Longrightarrow$ or $\Longleftrightarrow$

It is irrelevant whether one says $x-2=3 \Longleftrightarrow x=5$ or $x-2=3 \Longrightarrow x=5$ if we only want to go from $x-2=3$ to $x=5.$ Quite the opposite is the case: If someone writes $x-2=3 \Longleftrightarrow x=5$ when he only needs $x-2=3 \Longrightarrow x=5,$ then I would consider this as a bad style of writing since it makes the readers think about a statement, namely $x-2=3 \Longleftarrow x=5,$ where it is completely irrelevant. This is wasting other people's time. Very rude.

 

[SSequence]

Here are some of my thoughts on this topic. I have to say that I am not satisfied about my understanding of this (that is, looking at elementary equations from perspective of logic). There are number of things that can be confusing for me in this. Honestly, this is one of the reasons (amongst others ofc) that I feel to learn $PA$. I mean not just the syntax of it [and meaning of its statements] but also some specific deduction method related to fol. Even though $PA$ is restricted to natural numbers it might help me clarify my understanding of this better. But I haven't really gotten around to learning $PA$.

Regardless, here are some of my thoughts on this.

===============

First an answer to the question of OP. Let's restrict our domain to real numbers for the rest of this post. A statement like:
$\forall x \in \mathbb{R} \, \, [x-2=3 \implies x=5]$
is not incorrect. That's because the only way this can be false is when we can find a value of $x$ such that $x-2=3$ is true and $x=5$ is false. In other words, we have to find a value of $x$ other than $5$ such that $x-2=3$ is true. But such a value doesn't exist.

However, it is true that the following statements are also correct:
$\forall x \in \mathbb{R} \, \, [x-2=3 \impliedby x=5]$
$\forall x \in \mathbb{R} \, \, [x-2=3 \iff x=5]$
In particular, if you look at the first implication above it can be thought of as a definition of "solution" of the equation $x-2=3$.

===============

Generally speaking, we might think of an equation in one variable as being identified by a predicate $E:D \rightarrow \{0,1\}$. Here we have $D \subseteq \mathbb{R}$. We can categorize each value specific $r \in D$ either as a "solution" or as a "non-solution" of $E$. We have a specific value $r \in D$ as a solution iff $E(r)=1$. We have a specific value $r \in D$ as a non-solution iff $E(r)=0$. So the problem of solving an equation can be thought of as finding all the solutions. Let's denote $SS$ as the set of all solutions for $E$, which we can call the solution set of $E$.

In general for any $S \subseteq SS$ we will have:
$\forall x \in D \, \, [E(x) \impliedby x \in S]$
We will also have:
$\forall x \in D \, \, [E(x) \iff x \in SS]$

===============

Thee forward implication and backward implication both differ in the following ways:

(A) First let's take the forward implication. The forward implication means that we are retaining the complete solution set of the original equation $E$ in the latter equation step (while possibly including some non-solutions).

For the examples below, $x \in \{5,1\}$ is true exactly when $x$ takes on the values of $1$ or $5$ (and false for all other values of $x$). Informally I suppose one could also write it as $x=1,5$. Also, something like $x \in \{5\}$ can be taken as equivalent to $x=5$.

Here are some examples [$SS$ refers to the solution set of the equation on the left-hand-side]:
(A1) $\forall x \in \mathbb{R} \, \, [x-2=3 \implies x \in \{5\}]$
This is true as discussed before.

(A2) $\forall x \in \mathbb{R} \, \, [x-2=3 \implies x \in \{5,1\}]$
This is true as well. We have $SS=\{5\}$. We included a value on top of the solution set. But this doesn't alter the truth of the statement.

(A3) $\forall x \in \mathbb{R} \, \, [x^2=4 \implies x \in \{2,-2\}]$
This is true. We have $SS=\{2,-2\}$. Because all the values in the solution set are included it is true.

(A4) $\forall x \in \mathbb{R} \, \, [x^2=4 \implies x \in \{2,-2,1\}]$
This is true as well. We have $SS=\{2,-2\}$. We included a value on top of the solution set. But this doesn't alter the truth of the statement.

(A5) $\forall x \in \mathbb{R} \, \, [x^2=4 \implies x \in \{2\}]$
This is false. That's because the value $-2$ from the solution set wasn't included.

(A6) $\forall x \in \mathbb{R} \, \, [x^2=4 \implies x \in \{2,1\}]$
This is false as well. Once again, that's because the value $-2$ from the solution set wasn't included.

(A7) $\forall x \in \mathbb{R} \, \, [x^2=-4 \implies x \in \{1\}]$
This is true. The solution set is empty. Whether we add any values on top of it doesn't alter the truth of statement.

(B) Now let's take the backward implication. The backward implication means that we are retaining at least some values in the solution set of the original equation $E$ in the latter equation step (while not including any non-solutions).

Here are some examples for the same statements as before [$SS$ refers to the solution set of the equation on the left-hand-side]::
(B1) $\forall x \in \mathbb{R} \, \, [x-2=3 \impliedby x \in \{5\}]$
This is true since $\{5\} \subseteq SS$.

(B2) $\forall x \in \mathbb{R} \, \, [x-2=3 \impliedby x \in \{5,1\}]$
This is false. That's because a value $1 \notin SS$ has been added.

(B3) $\forall x \in \mathbb{R} \, \, [x^2=4 \impliedby x \in \{2,-2\}]$
This is true. That's because $\{2,-2\} \subseteq SS$.

(B4) $\forall x \in \mathbb{R} \, \, [x^2=4 \impliedby x \in \{2,-2,1\}]$
This is false. That's because a value $1 \notin SS$ has been added.

(B5) $\forall x \in \mathbb{R} \, \, [x^2=4 \impliedby x \in \{2\}]$
This is true. That's because $\{2\} \subseteq SS$.

(B6) $\forall x \in \mathbb{R} \, \, [x^2=4 \impliedby x \in \{2,1\}]$
This is false. That's because a value $1 \notin SS$ has been added.

(B7) $\forall x \in \mathbb{R} \, \, [x^2=-4 \impliedby x \in \{1\}]$
This is false. That's because a value $1 \notin SS$ has been added.

===============

Here is a somewhat generic way to think about it. Consider two equations (or statements) that can be identified by the predicates $E_1:\mathbb{R} \rightarrow \{0,1\}$ and $E_2:\mathbb{R} \rightarrow \{0,1\}$ [for simplicity, consider specific case of equations well-defined for all of $\mathbb{R}$]. Let $SS_1$, $SS_2$ denote the solution set of equations $E_1$, $E_2$ respectively. Then the statement:
$\forall x \in \mathbb{R} \, \, [\, E_1(x) \impliedby E_2(x) \,]$
is same as saying $SS_2 \subseteq SS_1$. Similarly the statement:
$\forall x \in \mathbb{R} \, \, [\, E_1(x) \implies E_2(x) \,]$
is same as saying $SS_1 \subseteq SS_2$.

As an example take $E_1$ to be $x^2=4$ and $E_2$ to be $x=2$. Then we have $SS_1=\{-2,2\}$ and $SS_2=\{2\}$.
If we take the statement:
$\forall x \in \mathbb{R} \, \, [\, x^2=4 \impliedby x=2 \,]$
The truth/falsity of the above statement is equivalent to $SS_2 \subseteq SS_1$. Since the latter is clearly true, the given statement is true.

If we take the statement:
$\forall x \in \mathbb{R} \, \, [\, x^2=4 \implies x=2 \,]$
The truth/falsity of the above statement is equivalent to $SS_1 \subseteq SS_2$. Since the latter is clearly false, the given statement is false as well.

 

[FactChecker]

The short answer to the OP is this: In a proof, include what is needed for the proof -- no more, no less.
If the proof needs a one-way implication, include the one-way implication.
The exception is when a proven theorem or lemma is invoked. Then use that "as is".

Any suggestion that using "$A \implies$ B" also means that $B \not\!\!\!\implies A$ is a false suggestion.

 

[Svein]

In predicate calculus $A\Rightarrow B$ is defined as $\neg (A \wedge \neg B)$, or (using deMorgan) $\neg A \vee B$. Expressed in words: You cannot have A and not B, or the converse: If A is false, B can be anything.