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We say that two statements are logically equivalent if and only if they have the same truth value in every model. I am wanting to be precise with this definition and was wondering if this is the same thing as the following: two sentences $\Gamma_1$ and $\Gamma_2$ are said to be logically equivalent, denoted $\Gamma_1\equiv \Gamma_2$, if and only if $\Gamma_1\models \Gamma_2$ and $\Gamma_2 \models \Gamma_1$. I am also curious is there a difference between saying a statement and a sentence here. Any help would be appreciated.

A sentence $\varphi$ is said to be a $\textbf{logical consequence}$ of a set of sentences $\Gamma$, denoted $\Gamma \models \varphi$, if and only if there does not exist a model $\mathcal{I}$ in which all members of $\Gamma$ are true and $\varphi$ is false.

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closed as unclear what you're asking by Rob Arthan, Shailesh, Xander Henderson, user91500, José Carlos Santos Sep 19 '17 at 20:26

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    $\begingroup$ Yes, those are the same conditions, but this is not obvious; it follows from the completeness theorem. $\endgroup$ – Qiaochu Yuan Sep 18 '17 at 22:34
  • $\begingroup$ That's what I needed. Thanks for your help. $\endgroup$ – W. G. Sep 18 '17 at 22:59
  • $\begingroup$ What do you mean by $\models$ here? (If $A$ and $B$ are sentences, most people write $A \models B$ to mean every model of $A$ is a model of $B$, so your question is trivial with that reading of $\models$.) $\endgroup$ – Rob Arthan Sep 18 '17 at 23:10
  • $\begingroup$ @QiaochuYuan: as the question appears to make no reference to proof, what has the completeness theorem got to do with it? $\endgroup$ – Rob Arthan Sep 18 '17 at 23:13
  • $\begingroup$ I am voting to close this pending clarification of the use of the symbol $\models$. $\endgroup$ – Rob Arthan Sep 18 '17 at 23:25
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Yes, these two definitions are equivalent. Here is a proof:

From 'left' to 'right':

Assume that in every model, $\Gamma_1$ and $\Gamma_2$ have the same truth-value. This means that there is no model where $\Gamma_1$ is true and $\Gamma_2$ is flase, and hence $\Gamma_1 \vDash \Gamma_2$. Likewise, there is no model where $\Gamma_2$ is true and $\Gamma_2$ is false, and hence $\Gamma_2 \vDash \Gamma_1$

From 'right' to 'left':

Assume $\Gamma_1 \vDash \Gamma_2$ and $\Gamma_2 \vDash \Gamma_1$. This means that there is no model where $\Gamma _1$ is true and $\Gamma_2$ is false, and there is also no model where $\Gamma _2$ is true and $\Gamma_1$ is false. So, there is no model where $\Gamma_1$ and $\Gamma_2$ have different truth_values, for if there were such a model, then in that model one of them would be true and the other false, and we just determined that there are no such models. Hence, in every model $\Gamma_1$ and $\Gamma_2$ have the same truth-value.

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  • $\begingroup$ Would you mind if I cited you as a source if I use any definitions/content we discuss on this site as well? I just wanted to let you know that I truly appreciate your help. I have given the answer to you, because it explains it more. Just to be particular, can it be the following where $\psi$ and $\varphi$ are formulas and are logically equivalent iff $\lbrace \varphi \rbrace \vDash \psi$ and $\lbrace \psi \rbrace \vDash \varphi$. I just worry about them being sentences here. $\endgroup$ – W. G. Sep 19 '17 at 19:41
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Yes, that is an equivalent definition.

This is assuming that every statement can be written as a sentence in formal logic and vice versa.

Typically, a statement is more general than a sentence.

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  • $\begingroup$ (I could be wrong.) $\endgroup$ – Shaun Sep 18 '17 at 22:33
  • $\begingroup$ Do you know of any difference between saying statement and sentence here? $\endgroup$ – W. G. Sep 18 '17 at 22:33
  • $\begingroup$ I've updated the answer, @W.G. $\endgroup$ – Shaun Sep 18 '17 at 22:35
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    $\begingroup$ @W.G.: The relevant difference between statement (or formula) and sentence is to do with the occurrence of free variables. But as you have accepted a vague answer and have not clarified an important lacuna in your question, I assume you are not interested in the details. $\endgroup$ – Rob Arthan Sep 18 '17 at 23:29
  • $\begingroup$ You're right, @RobArthan: my answer is a little vague. Would you mind answering it yourself, please? I'd like to see a better attempt . . . $\endgroup$ – Shaun Sep 19 '17 at 0:08

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