1

$$1) \alpha_1,\alpha_2,\alpha_3.......\alpha_{k-2}, \alpha_{k-1}, \alpha_k\vdash\alpha $$ Is a valid sequesnt. $$2) \alpha_1,\alpha_2,\alpha_3.......\alpha_{k-2}, \alpha_{k-1}, \alpha_k\models\alpha$$ $$3) \alpha_1\land\alpha_2\land\alpha_3.......\alpha_{k-2}\land \alpha_{k-1}\land \alpha_k\implies\alpha$$

From what I read in book . I am not being able to differentiate beteen these three arguments.Do they really mean same thing? If not,then when do they mean same thing?

Shubham Ugare
  • 703

2 Answers2

3

The "turnstile" : $\vdash$ means that $\alpha$ is derivable form the premises $\alpha_i$ according to the rules of the calculus (e.g. derivable from the premises and the logical axioms by way of modus ponens, or derivable with Natural Deduction).

The symbol : $\vDash$ means that $\alpha$ is a logical consequence of the $\alpha_i$, i.e. it is true whenever all the $\alpha_i$ are true.

The two relations are very different but stirctly linked : a calculus is said to be complete iff :

if $\Sigma \vDash \alpha$, then $\Sigma \vdash \alpha$.

In other words, a calculus is complete when it is "able" to derive from a set of premises all the logical consequences of the said premises.

(Quite) all the "interesting" calculus are complete.

For 3), I read $\Rightarrow$ as a synonim for $\to$, the conditional connective : "if ..., then ___".

In this case, the third relevant relation is established using the so-called Deduction (meta)-Theorem :

if $\alpha \vdash \beta$, then $\vdash \alpha \to \beta$.

As per your previous question, we can "iterate it" :

if $\alpha_1, \alpha_2 \vdash \beta$, then $\vdash (\alpha_1 \land \alpha_2) \to \beta$.

Mauro ALLEGRANZA
  • 94,169
  • +1, a good summary. What always helps me, but is (somewhat) nonstandard terminology, is to call $\vdash$ "provable consequence" and $\models$ "semantic consequence". Their finer relation becomes clearer when investigating beyond the "standard" logics and interpretations. – Lord_Farin Aug 17 '15 at 19:58
  • So does that mean there exist some "not complete " logical model in which a valid $\vdash$ doesn't imply $\models$.can you give example? – Shubham Ugare Aug 17 '15 at 20:24
  • 1
    No : the property : if $\Sigma \vdash \alpha$, then $\Sigma \vDash \alpha$ is called soundness : it means that the calculus is "correct", i.e. (roughly speaking) the calculus does not derive "false" formulae. Of course, we do not "like" unsound calculus. But the completeness property is more "interesting" : not all set of rules are complete, i.e. capable to derive all valid consequences. The completeness theorem for first-order predicate logic was proved in 1931 and was a remarkable mathematical result. – Mauro ALLEGRANZA Aug 17 '15 at 20:41
3

$\vdash$ is a symbol that refers to the rules of derivation of the system, whereas $\models$ refers to the semantics of the system. Hence, $\Gamma \vdash \alpha$ means that there exists a derivation (or, if you like, a proof) from $\Gamma$ to $\alpha$, whereas $\Gamma \models \alpha$ means that $\alpha$ is true in the model $\Gamma$ (or, if you like, that $\alpha$ is true given the assumption $\Gamma$).

Now, a desirable property of such systems is that $$\Gamma \vdash \alpha \iff \Gamma \models \alpha,$$ in which case the two are indeed "the same thing". $\Gamma \vdash \alpha \implies \Gamma \models \alpha$ is called soundness, whereas $\Gamma \models \alpha \implies \Gamma \vdash \alpha$ is called completeness. Whether or not you have a sound and complete logical system depends on the particulars of that system.

As far as your third example goes, if I'm being pedantic, it doesn't really mean anything, it is simply (presumably) a well-formed formula in your logic. That is, it is a string of symbols that can be constructed in your logic. You would need to prefix it with either a $\vdash$ or a $\models$ to make it clear whether you are talking about derivability or truth. However, especially when dealing with systems that are both sound and complete, we often dispense with the prefix.

mrp
  • 5,086