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I have a problem understanding the difference between $\to$, $\implies$ and $\models$.

The way I understand it is as follows:

$\to$ is used inside a formula between propositions. So saying $ A \to B $ is a formula.

$ \models $ is a statement about formulas. So saying $ A \land B \models A \lor B $ is making a statement about those two formulas $ A \land B $ and $ A \lor B $ and their relationship to each other.

And now comes my confusion: Where does $\implies$ come in all this?

My textbook made the following distinction:

"$ F \equiv \top $ is statement about formulas, so we can write (1) $F \equiv \top \implies G \equiv \top $. But this is not the same as (2) $F \models G$. "

But why not? Is it because $\models $ and $ \equiv$ are on the same level? And if yes, then on what level is $\implies$? Is it a statement about a statement about formulas?

Thank you

Mauro ALLEGRANZA
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Marwan
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1 Answers1

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Your book uses $\to$ as a logical connective (the conditional), i.e. as a "syntactical object" of the Language of propositional calculus (see page 15).

The symbol : $\top$ denotes the logical constant $\text {TRUE}$ and $\bot$ denotes the $\text {FALSE}$; they are not part of the language (in other cases, they are, and thus $F \to \bot$ is a formula, that can be used as definition for $\lnot F$).

The symbol $\equiv$ is used by your book not as the bi-conditional connective ($\leftrightarrow$, see page 16) but as a meta-mathematical abbreviation for the rleation of (logical) equivalence (see page 17).

Thus

$F \equiv \top$

is not a formula of the language but reads "the formula $F$ is equivalent to the True", i.e. "$F$ is always true", i.e. $F$ is a tautology (see page 20), and thus it is a statement about the formula $F$.

As explicitly stated in the book (page 19), the symbol is a synonym of the usual way to denote the fact that a formula $F$ is a tautology : $\vDash F$.

The book uses $\Leftrightarrow$ as an abreviation for "iff" : this symbol is not part of the language but is part of the meta-language, i.e. it is not used to write formulas but to express relations between formulas. In the same way, the book uses $\Rightarrow$ in the meta-language, to abbreviate "if..., then...".

Thus

$F≡ \top \Rightarrow G≡ \top$

is simply "if $F$ is a tautology, then $G$ is a tautology."

Also the symbol $\vDash$ is not part of the language of propositional calculus but is the meta-language symbol for logical consequence.

Thus

$F \vDash G$

reads : "there is no valuation $v$ such that $v(F)= \text T$ and $v(G)= \text F$," or (more plainly) :

"$G$ is TRUE in every interpretation where $F$ is TRUE."

The two relations are not the same : we have that "$F \vDash G$" is stronger than "if $F$ is a tautology, then $G$ is a tautology".

Consider the case with $P \lor Q$ as $F$ and $Q$ as $G$; clearly $P \lor Q \nvDash Q$, but $P \lor Q$ is not a tautology, and thus :

if $P \lor Q$ is a tautology, then $Q$ is a tautology,

holds.

Note : compare with van Daleen's textbook, where $\to$ and $\leftrightarrow$ are the conenctives while $\Rightarrow$ and $\Leftrightarrow$ are used as abbreviations in the meta-language

Mauro ALLEGRANZA
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  • so $\implies$ and $\models$ are both on the same level (statement about formulas), but mean different things? And $\equiv$ and $\to$ are also on the same level, inside a formula? – Marwan Jul 05 '19 at 12:34
  • Ok i think i understand that now. But one last question: how does $ P \lor Q \equiv \top \implies Q $? Q could be false in the first formula and that would still make the first formula true, but then Q wouldn't be true. Or did i miss something? – Marwan Jul 06 '19 at 08:55
  • @MarwanEzzat - I've used $P,Q$ as propositional variablles (like $A,B$ in your textbook); thus $P \lor Q$ is not a tautology. – Mauro ALLEGRANZA Jul 06 '19 at 10:48