Open axioms of equality

Solution 1:

Yes, we can.

Consider the Natural Deduction proof system.

We can derive $(x=x)$ from the "closed" version: $\forall x \ (x=x)$ simply using $\forall$-elim rule.

Conversly, from the open axiom $\vdash (x=x)$, we can derive $\vdash \forall x \ (x=x)$ using $\forall$-intro.

There are no assumptions (i.e. $\Gamma = \emptyset$) and thus the proviso of the rule: "$x$ not free in $\Gamma$", is satisfied.

The open equality axiom $(x=x)$ is obviously valid.

See e.g.:

• Dirk van Dalen, Logic and Structure, Springer (5th ed. 2013), page 67.

The definition of true in a structure $\mathfrak A$ is restricted to sentences, i.e. "closed" formulas.

For open ones, it adopts the convention that:

$\mathfrak A \vDash \varphi \ $ iff $ \ \mathfrak A \vDash \text {Cl} (\varphi)$,

where $\text {Cl} (\varphi)$ is the universal closure of $\varphi$.

A different approach is adopted by:

• Herbert Enderton, A Mathematical Introduction to Logic, Academic Press (2nd ed. 2001), page 83.

In this case, the meaning and the truth-value of an open formula with respect to an interpretation $\mathfrak A$ is defined for specific "instances" of the formula, obtained through the variable assignment device.

The basic notion is that of satisfaction of a formula $\varphi$ by an assignment $s$ in a interpretation $\mathfrak A$ (in symbols: $\mathfrak A, s \vDash \varphi$).

A formula $\varphi$ is true in $\mathfrak A$ when it is satisfied by every assigment.

Consequently, $(x=x)$ is true in every interpretation.