Substitution of identicals in predicate logic

Question

Given a variable $x$ and a variable $y$, is it correct to state that $P(x)$ is equivalent to $P(y) \land x = y$? That the latter implies the former seems clear to me due to the rule of substitution. But I still have problems trying to prove the contrapositive.

Answer 1

They are not equivalent.

Think semantically: two formulas $\varphi$ and $\psi$ are equivalent iff in every structure $\mathcal{A}$, every assignment of variables to elements of $\mathcal{A}$ makes $\varphi$ and $\psi$ have the same truth value.

But consider a structure with two distinct elements $a$ and $b$ such that $P(a)$ and $P(b)$ both hold. Then the variable assignment sending $x$ to $a$ and $y$ to $b$ makes your first formula true but your second formula false. You might object that this assignment is ruled out by your second formula, but what's actually happening is that this assignment is making the second formula false, which is fine (we have to consider all possibilities).

---

It may help to first consider the universal closures of the formulas involved: "$\forall x(P(x))$" and "$\forall x,y(P(y)\wedge x=y)$" are clearly not equivalent (the latter says additionally that the universe has only one element).

• Incidentally, equivalence of universal closures does not imply equivalence: consider e.g. $$x=y\wedge\exists u,v(u\not=v)\quad\mbox{ and }\quad x\not=y$$ versus their universal closures $$\forall x,y(x=y\wedge\exists u,v(u\not=v))\quad\mbox{ and }\quad\forall x,y(x\not=y).$$ The universal closures are equivalent since they're each never true, but the original formulas are not equivalent since there are some structures with some variable assignments giving them different truth values (e.g. look at any more-than-one-element structure and consider the valuation sending $x$ and $y$ to different elements).