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I have several questions about entailment and implication.

1) Let $A = \{p_1, ..., p_n \}$ be some set of logical statements and $q$ some logical statement. Is it right that "$A \models q$" is a correct logical statement, i.e. it may be true, or false and we may always define its logical meaning?

2) What if $A = \varnothing$?

3) Is it right that $(A \models p) \equiv ((p_1 \land ... \land p_n) \rightarrow q)$ ?

4) How should i treat the following situation: $\{ q, q->p \} \models p \rightarrow r$. Is it a correct entailment? I'm confused because there is $r$ on the right and there is no $r$ on the left.

Thanks in advance.

EDIT

Sorry for my English. I suppose "logical proposition" is a more correct term then "logical statement".

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  • If $q$ is a "logical statement" (assuming you mean is logically valid), then we have $\vDash q$, and so a fortiori $A \vDash q$. (If $q$ is true on all interpretations, it is true in particular on all intepretations which make the members of $A$ true.)

  • If $q$ is a "logical statement", then we have $\emptyset \vDash q$. (For $\emptyset \vDash q$ says that $q$ is true on those interpretations which make any $p$ true if $p$ is a member of $\emptyset$ -- which are, vacuously, all interpretations).

  • Not quite. You mean $(A \models q)$ iff $\models ((p_1 \land ... \land p_n) \rightarrow q)$ or $(A \models q) \Leftrightarrow\ \models ((p_1 \land ... \land p_n) \rightarrow q)$ [You want metalinguistic claims on both sides of the equivalence.]

  • Finally $\{ q, q->p \} \models p \rightarrow r$ is plainly not correct. Suppose $p$ and $q$ are true, and $r$ false.

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  • $\begingroup$ You seem to be using "logical statement" for what I would call "logically valid statement". Is this standard terminology, or a typo? $\endgroup$ – Henning Makholm Mar 22 '14 at 12:29
  • $\begingroup$ Well, I was using quotes to signal I was indeed quoting the OP, and I took him indeed to mean logically valid statement -- I've clarified! $\endgroup$ – Peter Smith Mar 22 '14 at 12:48
  • $\begingroup$ Sorry, I was confused because the question seemed to make better sense to me if "logical statement" meant "wff" and "correct" means "well-formed" rather than "true". $\endgroup$ – Henning Makholm Mar 22 '14 at 13:09
  • $\begingroup$ Well I agree it wasn't clear ...but I though he was worrying about cases where we say that a [logically valid] $q$ is entailed by a load of irrelevant $p_i$, or by nothing. $\endgroup$ – Peter Smith Mar 22 '14 at 13:17
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I'm assuming (contrary to Peter Smith), that by "logical statement" you mean "well-formed formula" or possibly even "sentence".

1) It is true that if $p_1$ through $p_n$ and $q$ are formulas, then we can say $p_1,\ldots,p_n\vDash q$, and that $p_1,\ldots,p_n\vDash q$ will be either true or false. But $p_1,\ldots,p_n\vDash q$ is not itself a formula. It is a claim about formulas at the metalevel.

A metalevel claim such as $\vDash q$ has a definite truth value in itself, (though it can be difficult to determine what it is). On the other hand a formula only acquires a truth value when we specify an interpretation for the non-logical symbols in it.

2) All of the above holds for empty sets of assumptions too.

3) No, because $A\vDash p$ is a property at the metalevel and $(p_1\land \cdots \land p_n)\to q$ is a formula itself. They are not even the same kind of things, so they can't be equivalent.

4) $p, p\to q\vDash p\to r$ is a meaningful claim, which just happens to be false. I suspect you're using "correct" to mean "meaningful" rather than "true", in which case you would probably call it "correct".

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