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I've heard of all sorts of logic: Mathematical logic, propositional logic, first order logic, second order logic, predicate logic, Boolean logic / algebra, etc.

What are the main differences between these categories and how do I know what to apply when?

user525966
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1 Answers1

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To give a very good answer would require probably to read a good book in mathematical logic.

Nevertheless here is a very short, and possibly rough, description of the terms you asked for. Be careful, what follows is a very imprecise description of the various kind of logic, but again I think this is the best one can get in the limit of a post (or at least, that is the best I can think of).

Mathematical logic denote that branch of mathematics that applies the technique of mathematics to the study of logics. It basically provides formal ways to define what a logical system is and how to prove stuff about these systems.

Propositional, first-order and second order logics are just few examples of these formal logical systems.

The differences in these systems are in the expressiveness of the said systems.

For instance in propositional logic one deals only with true-or-false statements while in first-order logic one can use formulas which states that certain relations hold for some objects or for all the objects.

In second-order logic we can start to state that relations between relations hold for certain relations or for all the possible relations, something which cannot be done in first-order logic.

The term Boolean logic is usually used a synonym for propositional logic.

Boolean algebra is a term used for denote a certain class of algebraic structure which captures the algebraic properties of propositional logic.

Edit: by the way if you look at wikipedia's article you can find lots of informations on the subjects.

Edit 2: the OP asked in the comments below for an example of second order sentence, here is an informal example.

In what follow I will use the following notation: $P x$ will mean $x$ satisfies the property (unary relation) $P$ and $x R y$ will mean $x$ is in the relation $R$ with $y$.

For every binary relation $R$ such that

  • for all $x$ we have that $x R x$
  • forall $x$ and $y$ if $x R y$ and $y R x$ then $x=y$
  • forall $x$, $y$ and $z$ if $x R y$ and $y R z$ then $x R z$
  • forall properties $P$ such that exists at least an $x$ for which $P x$ there is a $m$ such that $P m$ holds and for every other $x$ for which $Px$ holds we have $m R x$

we have that forall $x$ and $y$ either $x R y$ or $y R x$.

This apparently complex statement is basically saying that every order $R$ which is wellorder, i.e. an order such that any satisfied property $P$ has a minimal element that satisfies it, is a total order.

MJD
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Giorgio Mossa
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  • "In second-order logic we can start to state that relations between relations hold for certain relations or for all the possible relations, something which cannot be done in first-order logic." w-w-...what? XD Pardon my ignorance; is there a good/easy example of what you mean by this? – user525966 Aug 30 '18 at 16:16
  • @user525966 sure, if you want I can add the example to the answer (trying to stay the as informal as possible). – Giorgio Mossa Aug 30 '18 at 16:32
  • @user525966 done, does the the edit2 satisfy you? – Giorgio Mossa Aug 30 '18 at 16:46
  • Yes, thank you. In practice do people need to get into second order logic as much? It seems a lot more involved – user525966 Aug 30 '18 at 18:43
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    @user525966 well I would say no but that is because people usually do mathematics in a set theory. In this way they can represent properties and relations as sets, which becomes object of the theory, and so they can do everything using just a first-order logic. But if you are interested in studying relations (or even relations between relations and so on) outside of a theory of this kind you need an second (or even higher order logic). – Giorgio Mossa Aug 30 '18 at 20:05
  • So it seems like as we increase the "order" to order n, we just start talking about ways to fiddle with the rules/quantifications of the "n-1"th order logic? – user525966 Aug 30 '18 at 20:06
  • @user525966 Sort of. – Giorgio Mossa Aug 30 '18 at 20:57