Equivalence of Definitions of Semantic Equivalence for Predicate Logic

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Theorem

Let $\mathbf A, \mathbf B$ be WFFs of predicate logic.


The following definitions of the concept of Semantic Equivalence (Predicate Logic) are equivalent:

Definition 1

Then $\mathbf A$ and $\mathbf B$ are equivalent if and only if:

$\mathbf A \models_{\mathrm{PL_A}} \mathbf B$ and $\mathbf B \models_{\mathrm{PL_A}} \mathbf A$

that is, each is a semantic consequence of the other.


That is to say, $\mathbf A$ and $\mathbf B$ are equivalent if and only if, for all structures $\AA$ and assignments $\sigma$:

$\AA, \sigma \models_{\mathrm{PL_A}} \mathbf A$ if and only if $\AA, \sigma \models_{\mathrm{PL_A}} \mathbf B$

where $\models_{\mathrm{PL_A}}$ denotes the models relation.


Definition 2

Then $\mathbf A$ and $\mathbf B$ are equivalent if and only if:

$\mathbf A \iff \mathbf B$ is a tautology

where $\iff$ is the biconditional connective.


Proof

Let $\AA$ be a structure for predicate logic.

Let $\sigma$ be an assignment for $\mathbf A \iff \mathbf B$ in $\AA$.


Then the value of $\mathbf A \iff \mathbf B$ under $\sigma$ is given by:

$\map {f^\leftrightarrow} {\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma, \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma}$

and from the definition of $f^\leftrightarrow$ we see that $\AA, \sigma \models_{\mathrm{PL_A}} \mathbf A \iff \mathbf B$ if and only if:

$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma$

Because the possible values are just $\T$ and $\F$, this is equivalent to:

$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \T$ if and only if $\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma = \T$

which by definition of $\mathrm{PL_A}$-model amounts to:

$\AA, \sigma \models_{\mathrm{PL_A}} \mathbf A$ if and only if $\AA, \sigma \models_{\mathrm{PL_A}} \mathbf B$


Because $\AA$ and $\sigma$ were arbitrary, the above equivalence holds for all such $\AA$ and $\sigma$.

The result follows by definition of tautology.

$\blacksquare$