Can anyone give a simpler definition of The Truth Schema?

Question

The definition given in our material is quite long and hard to understand. Can anyone rephrase it or give a simpler definition?

Answer 1

The formal definition can hardly be shortened, but maybe I can make you better understand it by giving an example:

Suppose we use the following predicate symbols: $Square$ and $Larger$. $Square$ takes 1 argument,and $Larger$ takes two. Suppose also that we have 2 constant symbols $a$ and $b$. So, now we can create sentences like:

$a=b$ (an atomic sentence)

$Square(a)$ (another atomic sentence)

$Larger(a,b)$ (yet another atomic sentence)

$\neg Square(b)$ (a complex sentence involving truth-functional operator(s))

$\forall x \: Square(x)$ (a complex sentence involving a quantifier)

OK, these are all just statements though and, like any statement, they can be true or false. How do we know if they are true or false? Well, that depends on what kind of a world (or structure) they are evaluated in, and how those sentences make claims about those worlds. In other words, we need a structure, or interpretation. In the formal definition, this is $\mathcal{M}$. The structure defines:

1. A domain of discourse: what set of objects exist in the world that we want the sentences to be about. In the formal definition, this is $M$

2. An interpretation for each of the 'non-logical' symbols ... so those are the predicate symbols like $Square$ or constant symbols like $a$

So, for example, let us consider a world with 3 objects, and call them $o_1$,$o_2$, and $o_3$. That is: $M = \{ o_1,o_2,o_3 \}$

Let us say that we want $Square(x)$ to mean 'x is a square', and let's say that $o_1$ and $o_2$ are squares, while $o_3$ is not. In the formal definition, we define this as: $Square^\mathcal{M} = \{ o_1, o_2 \}$

Let us say that we want $Larger(x,y)$ to mean 'x is larger than y', and let's say that $o_3$ is larger than $o_1$ and $o_2$, but $o_1$ and $o_2$ are the same size. In the formal definition, we define this as: $Larger^\mathcal{M} = \{ (o_3, o_1),(o_3, o_1) \}$

Finally, let's say that both constant symbols $a$ and $b$ refer to object $o_1$ (this is of course always possible: the same object can go by multiple names). In the formal definition, this would be $a^\mathcal{M} = o_1$ and $b^\mathcal{M} = o_1$

OK, so now we can finally determine the truth of the statements:

$a=b$: This is true, since $a$ and $b$ denote the same object $o_1$. Formally, we write $\mathcal{M} \vDash a = b$ (that is: the sentence $a = b$ is true under the structure $\mathcal{M}$) since $a^\mathcal{M} = b^\mathcal{M}$

$Square(a)$: This is true, since $a$ denotes $o_1$, and $o_1$ is a square. Formally: $\mathcal{M} \vDash Square(a)$ (which can also be expressed as '$\mathcal{M}$ models sentence $Square(a)$) because $a^\mathcal{M} \in Square^\mathcal{M}$.

$Larger(a,b)$: This is not true since $(a^\mathcal{M},b^\mathcal{M}) \not \in Larger^\mathcal{M}$. (that is: $(o_1,o_1) \not \in \{ (o_3, o_1),(o_3, o_1) \}$)

$\neg Square(b)$: Since we have $\mathcal{M} \vDash Square(b)$ (since $b^\mathcal{M} = o_1 \in Square^\mathcal{M}$), we have not $\mathcal{M} \vDash \neg Square(b)$. So this statement is false under this structure, also written as $\mathcal{M} \not \vDash \neg Square(b)$

$\forall x \: Square(x)$ : This statement would be true iff all objects in M are squares, i.e. for every $o \in M$: $o \in Square^\mathcal{M}$. That is not the case (since $o_3 \not \in Square^\mathcal{M}$), so this statement is false.

OK, so I have given a concrete example that hopefully makes more sense, and I have indicated the connections as to how the formal/mathematical definition tries to express these basic ideas. The one complication I did avoid is that you can be dealing with complex terms when you have function symbols in your language, in which case you need to define what those function symbols mean in your structure. But, first try and get a good grasp of this, and then you can move on to function symbols and complex terms if needed. Good luck!

Answer 2

It's basically saying that the meaning of $\models$ is defined by substitution.

For example,

$$(\mathbb{R},+ \mapsto \max,\times \mapsto +) \models \forall x(x+x=x) \iff \forall x \in \mathbb{R}(\max\{x,x\}=x)$$

So you go through and replace every copy of the symbol $+$ with the function $\max$, every copy of the symbol $\times$ with the function $+$, etc. and you wind up with the meaning of $(\mathbb{R},+ \mapsto \max,\times \mapsto +) \models \varphi.$

Makey sensey?

Answer 3

I think the issue you have here is that the given material does not explain the fact that one must work in an existing system, normally called the meta-system, in which one already has at least the ability to perform conditional operations, so that one can define a boolean connective (in the target system) via a truth-table, which is basically many conditional rules such as "if $p=0$ and $q=1$ then $p \land q \oversetΔ=0$". One almost always assumes classical logic in the meta-system.

For example we want to define the meaning of "$A \lor B$" in the target system $S$. If we want $S$ to have classical logic, and want to use "$\lor$" for disjuction, then naturally we would define "$A \lor B$" to be true if "$A$" is true or "$B$" is true, and false otherwise. As said earlier, this can be captured by a truth-table. Similarly for all the other boolean connectives.

Truth itself is relative to a structure (world). So instead of just saying "true/false" we say "true/false in $M$", where $M$ is a structure. Of course $M$ has to specify the interpretation (meaning) of atomic sentences (in propositional logic) or function-symbols and predicate-symbols (in first-order logic). And we write "$M \vDash A$" to mean "$M$ satisfies $A$" or "$A$ is true in $M$". It could be that we have two structures $M,N$ such that $M \vDash A$ but $N \nvDash A$ for some sentence $A$. But the whole point of classical logic is that the interpretation of a compound sentence in any structure is completely determined by the interpretation of the constituent parts in that structure.

Using the same example, we choose to define (in the meta-system) for any first-order structure $M$ and any sentences $A,B$, that ( $M \vDash A \lor B$ ) iff either ( $M \vDash A$ ) or ( $M \vDash B$ ). Essentially this makes the 'internal' disjunction match the 'external' (meta) disjunction, which is what we want; the goal of logic was to use syntax (symbols) to pin down the intended semantics (meaning) as far as possible. Notice that one must define the interpretation of sentences for all the notations at once, because a compound sentence may use all the possible notations. One must also verify that such a definition is well-founded, and to do so we in fact have to prove a theorem that there is only one way to parse each first-order sentence, namely to determine whether it is an atomic sentence or a compound sentence and if so of what type and how it breaks down.

Note that the semantics of quantifiers cannot be captured by a truth table. Given any structure $M$ and any sentence $A$ and any variable $x$, we define that ( $M \vDash \forall x\ (A)$ ) iff ( $M [x:=c] \vDash A$ for every element $c$ in $M$ ). Here "$M [x:=c]$" is defined to mean a structure that is the same as $M$ except that for any sentence it first interprets any free occurrence of "$x$" as $c$, before using $M$ to interpret the rest of the sentence. As before, we are essentially defining the 'internal' quantifier to match the 'external' quantifier. [Note that this is where classical logic gets non-computable; in the meta-system we assume that the quantified statement "... for every element $c$ in $M$" is either true or false, but even if it is so we have no way of knowing which case it is. In contrast the boolean operations can be computed easily.]

An important point is that for the above recursive definition to work we cannot just define the interpretation of sentences, but must define the interpretation of general formulae, which may have free variables. There are multiple incompatible ways to do this. One way (that ties in with Hilbert-style deductive systems) is to stipulate that every structure must already interpret every variable as some element. So for example a structure $M$ will already interpret "$x$", but of course "$M [x:=c]$" will interpret "$x$" as $c$ instead of whatever $M$ interprets it as. Another way is to only allow writing "$M \vDash φ$" (in the meta-system) when $M$ interprets all the free variables of "$φ$". Both of these ways are consistent with the definition I gave above for the semantics of quantifiers, but I have not seen the latter way in textbooks.