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.
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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##.
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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]##
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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.
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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.