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Imagine that you know that a given fact holds for every value of x. Then it must surely hold after you substitute some particular value for x. For example, you know that x + 1 = 1 + x for every x. So it must be the case that 3 + 1 = 1 + 3. The latter equation is a specialization of the former equation.
You are not limited to replacing x by a constant. You can replace it by any expression. For example, another specialization of equation x + 1 = 1 + x is (uv + 2) + 1 = 1 + (uv + 2). For specialization, be sure to do the substitution for every occurrence of the variable.
What if you have a statement that is true for all values of several variables? For example, you know that x + y = y + x for all x and y. Then you can replace each of the variables by any expression at all, and you get another true statement. So (u+1) + 2z = 2z + (u+1).
This all probably sounds very obvious. Yet, it represents a constant difficulty. Many of the facts that we will encounter are unfamiliar. But they work the same way. Suppose that you know that fact P(f(x), g(x)) holds for every x. You do not need to know what it means. Your goal is to prove that P(f(h(y)), g(h(y))) is true. That is no problem. Replacing every occurrence of x by h(y) in P(f(x), g(x)) yields exactly what you want to prove.