Is $2$ a substitution for $1+1$, or equal to $1+1$? In this case both seems true, but I was wondering if there really is a difference. Whenever there is an equality $A=B$, is it also true that $B$ is always a substitution for $A$?
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$\begingroup$It is true that if there's equality, then you can apply substitution. That comes from the definition of equality, as explained in this answer. The opposite is not always true. Let's see some examples.
Non-transitive relations
Here, substitution doesn't work because there's no equivalence relation at all. If you take two real numbers $x$ and $y$, given that $x \ne y$ and $x \ne 7$, it would be a huge mistake to claim that $y \ne 7$.
Transitive relations
Let's add the transitivity. The subset inclusion relation $\subset$ suits well. Knowing, for example, that
$$ \lbrace x \in \mathbb{R} \colon -1 \le x \le 1 \rbrace \subset \lbrace x \in \mathbb{R} \colon 1 + \lvert x \rvert < \pi \rbrace $$
and that $\mathbb{Q} \cap [0,\frac12]$ is included in the first set, you can safely write that
$$ \mathbb{Q} \cap [0,\frac12] \subset \lbrace x \in \mathbb{R} \colon 1 + \lvert x \rvert < \pi \rbrace $$
but there's no equality in those sets. Indeed, you would have that
$$ \lbrace x \in \mathbb{R} \colon -1 \le x \le 1 \rbrace \subset \mathbb{Q} \cap [0,\frac12] $$
which is false.
Substitution derived from equality
That's the case of, say, algebraic substitution. Suppose you are given the system of equations
$$ \begin{cases} x + y = 6 \\ 3x + 2y = -1 \end{cases}; $$
you can rephrase the second equation as
$$ 2(x+y) + x = -1 $$
and substitute $x + y$ with $6$ because of their equality, leading to
$$ \begin{cases} x = -13 \\ y = 19 \end{cases}. $$
The peculiar feature of equality is that, among the axioms which define it, there is the so-called indiscernibility of identicals.
Given two entities $x$ and $y$ and a predicate $\phi(u, \alpha_1,\dots,\alpha_n)$, $$ x = y \implies \phi(x,\alpha_1,\dots,\alpha_n) = \phi(y,\alpha_1,\dots,\alpha_n). $$
Can you recognise which entities are involved in this example? Which is the predicate?
Then, there is Leibniz's Law, which states that also the opposite implication holds for equality, giving the equivalence
$$ \forall x.\forall y. x = y \iff \phi(x,\alpha_1,\dots,\alpha_n) = \phi(y,\alpha_1,\dots,\alpha_n). $$
Substitution without equality
If you take the relation
$$ x \equiv 3 \pmod{6} \quad x \in \mathbb{Z} $$
and thanks to the transitivity of the congruence relation, we can substitute $3$ by $9$, since they are congruent modulo $6$. However, they are not the same integer. As you can see, here substitution is used without the condition of equality.
If you take the set of congruence classes modulo $6$, denoted by $\mathbb{Z}/6\mathbb{Z}$, you have that $[3]$ and $[9]$ are equal. That's a matter of definition of the involved entities.
What about quotients?
As nicely explained here, a necessary condition to have equality and substitutivity is for the considered relation to be compatible with operations. Then, the quotient space w.r.t. the equivalence relation allows you to consider compatible objects as equals.
$\endgroup$ 6 $\begingroup$In the sense you have defined substitution and equality, they refer to the same thing.
Firstly, in order to shorten long complicated expressions containing $1+1$, we chose to substitute it by $2$. Hence we made the statement that $1+1=2$.
However, one can also claim that we have DEFINED $1+1=2$.
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