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Most of the time, the equivalence between two formulas is shown when they are just a refactoring form of each other, e.g., is it equal

$(p \rightarrow q) \equiv (\neg p \vee q)$

In this example, both formulas have the same variable names.

My question is: if

$ \phi_a = (m \wedge p) \vee (m \wedge q)$ ,

$\phi_b = (x \vee y) \wedge z$ ,

knowing that $\phi_c = (p \leftrightarrow x) \wedge (q \leftrightarrow y)\wedge (m \leftrightarrow z)$

How to cheeck whether $\phi_a \equiv \phi_b$ ?

One way that I am trying is to substitute the variables from $\phi_c$ to either $\phi_a$ or to $\phi_b$. But, does it make sense if I try simply $\phi_a \wedge \phi_b \wedge \phi_c$ ? When I am trying this in a solver, it is generating the same set of models with $\phi_a$ and $\phi_b$, just twice longer (including the variables from both formulas).

So, am I acting correctly? If not, what I should do?

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    @Mauro ALLEGRANZA , yes it is, but this is a small example and I want to avoid trying to refactor formulas for making them equivalent, plus still the `m`, `p`, and `q` I have to relate with $\phi_c$ links to `x`, `y`, and `z`, because both formulas can have different meaning although the same syntax. So, how to involve these links in both formulas? – 2017-02-03
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    [Replacement Theorem](https://books.google.it/books?id=q-YEuuZ_j8EC&pg=PA122). – 2017-02-03

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Yes, you have the right idea. The relevant equivalence is:

Substitution Equivalence

Where $S(\phi)$ is some formula that contains formula $\phi$ as a component part (possibly multiple times), and where $S(\psi)$ is the sentence that results from replacing any of the formulas $\phi$ with formula $\psi$, then:

$S(\phi) \land (\phi \leftrightarrow \psi) \Leftrightarrow S(\psi) \land (\phi \leftrightarrow \psi)$

So in your case, you can start with $\phi_a \land \phi_c$ and then proceed as follows:

$\phi_a \land \phi_c =$

$((m \land p) \lor (m \land q)) \land ((p \leftrightarrow x) \land (q \leftrightarrow y) \land (m \leftrightarrow z)) \Leftrightarrow$

$((m \land (p \lor q)) \land ((p \leftrightarrow x) \land (q \leftrightarrow y) \land (m \leftrightarrow z)) \Leftrightarrow$

$(m \land (p \lor q)) \land (p \leftrightarrow x) \land (q \leftrightarrow y) \land (m \leftrightarrow z) \Leftrightarrow$ (and now you use it!)

$(z \land (x \lor y)) \land (p \leftrightarrow x) \land (q \leftrightarrow y) \land (m \leftrightarrow z) \Leftrightarrow$

$((x \lor y) \land z) \land ((p \leftrightarrow x) \land (q \leftrightarrow y) \land (m \leftrightarrow z)) =$

$\phi_b \land \phi_c$

So note: it is not the case that$ \phi_a \Leftrightarrow \phi_b$, but you do have that $\phi_a \land \phi_c \Leftrightarrow \phi_b \land \phi_c$

So you can't say that $\phi_a$ and $\phi_b$ are equivalent, but you can say that within the context of $\phi_c$, $\phi_a$ and $\phi_b$ say the same thing.

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    I was just reading about the `Replacement Theorem` that @Mauro ALLEGRANZA suggested. You have no idea how much I am happy that you showed this prove. And, it is exactly about what I was asking. Also, thanks for the last sentence; it is very helpful to know how to be articulated in the right way in this case. Thank you again, Sir/Madam :). – 2017-02-03
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    @user4712458 You're welcome! The replacement theorem is very much the same idea, yes, though a little harder to directly apply to your case then this substitution equivalence. – 2017-02-03