Problems with using validity symbol ⊨ "vacuously", as in "X ⊨" and "⊨ A"

Question

Something has been bothering me for a while now. Sorry if the length doesn't seem to warrant the subject matter but it's been driving me nuts.

In Greg Restall's Logic An Introduction, p. 56 to be precise, some claims are made about ways to use the ⊨ symbol. The first one is "⊨ A iff any evaluation satisfies A." The second one is that "X ⊨ iff there is no evaluation satisfying everything in X."

I'll first quote how ⊨ is defined in this book.

Let's first summarize the definition: X ⊨ A iff any evaluation satisfying everything in X also satisfies A. Or equivalently, if you prefer a negative statement, we have X ⊨ A iff there is no evaluation satisfying everything in X that doesn't satisfy A."

The author then continues with the explanation of the 2 claims mentioned above:

"Now, this works for any collection of formulas. In particular, it works when X is empty. We write this as '⊨ A'. What does this mean? Well, it means that any evaluation satisfying everything in the empty set also satisfies A. Well, satisfying everything in the empty set is really easy. (After all, there is nothing in the empty set for an evaluation to make false!) So '⊨ A' has this definition: ⊨ A iff any evaluation satisfies A. So, ⊨ A iff A is a tautology."

And the second one:

"Another one is we leave out the formula A. here, X ⊨ holds iff there's no evaluation satisfying everything in X that doesn't satisfy... what? There's nothing to satisfy in the conclusion. So, we have this definition: X ⊨ iff there is no evaluation satisfying everything in X. So, X ⊨ iff the set X is unsatisfiable."

I really have big issues with how all of this is supposed to hold. For instance, what about "(p v ~p) ⊨"? The book claims it's "really easy" to satisfy everything in the empty set. Then surely any valuation satisfying everything in X also satisfies everything in the empty set: ones in which p is true as well as ones in which p is false. The "easy to satisfy" empty set should be satisfied with either. But Restall defined X ⊨ A as just when any valuation satisfying everything in X also satisfies A. So I fail to see how X ⊨ is supposed to be the same as saying X is false, since (p v ~p) is of course always true. The same seems to hold for X being any satisfiable formula, not just tautologies and not just for formulas that might be suspect due to intuitionist notions or whatever.

It seems wrong to in one instance say of the empty set that "there is nothing to satisfy" as if this implies that no valuation whatsoever satisfies it but then to also claim that it's "easy to satisfy" as if to imply that every valuation whatsoever satisfies it. The problem to me seems to be that two different standards are used to evaluate the satisfiability of an empty set.

In the case of X, what comes before the ⊨, a universal approach seems to be taken: for all that is in the empty X it is true that it is satisfied by whatever valuation. But for A, what is to the right of ⊨, an existential one seems to be used: there is no formula in there that is satisfied by whatever valuation.

I just fail to see what justifies these different approaches to X and A if that is at all the issue here. I mean, the definition does say "satisfying everything in X" but why couldn't it have worked the same way if it said "satisfying X", and likewise "satisfying everything in A" rather than "satisfying A"? Everything would work the same way to the best of my insight, except the vacuous examples, but that could as well be totally arbitrary, a mere choice, as far as I can see.

Of course there is an asymmetry between X and A because X ⊨ A isn't the same as A ⊨ X. And yes, this seems to link universal quantification with X and existential with A, but this with respect to valuations not formulas: every valuation that makes the entire of X true must also make A true, while some of the valuations that make A true must be the only valuations making X true. However I fail to see how this justifies treating an empty X differently in that regard from an empty A because that concerns (the lack of) formulas not valuations.

The best I can come up with involves the fact that for satisfaction purposes, each set of formulas can be transformed to an equivalent formula which is basically the disjunction of, for every valuation satisfying the set, the conjunction of, for each atomic proposition involved, the negation or simply that proposition, whatever the respective valuation dictates. For example (p v q) becoming (p and ~q) or (~p and q) or (p and q). Then, if we take an empty initial set with no formulas to transform, well there is nothing to transform so we end up with a list of "no valuations" (there will be none of the disjuncts that correspond to a valuation). Then every valuation to be found in the list would also make A true in the case of ⊨ A. Conversely, there would be no valuations to be had from the empty set A that will be the only ones to satisfy X. But in the case of ⊨ A that doesn't get us where we want because we don't want every valuation to be found in the empty list, which are none, we want all of them.

So to conclude, I can't really put my finger on what I'm really missing here or otherwise - on the off chance - how I'm supposed to demonstrate that what Restall is doing is definitely not right. In the case that it's indeed related to the universal vs existential quantification thing I mentioned, I do not really see what justifies it or how it explains away the counterexamples you can make by having X filled with a single satisfiable or even tautological formula.

Any ideas?

Answer 1

Restall seems to use a generalized notion of consequence (due to Gentzen) which holds between SETS of formulas. Let $L$ be some standard first-order language with the usual satisfaction relation. Then for any sets $\Gamma, \Delta$ of $L$-sentences let $\Gamma \models \Delta$ iff for every $L$-model $\mathcal{M}$: If $\mathcal{M}$ satisfies every member of $\Gamma$, then $\mathcal{M}$ satisfies at least one member of $\Delta$. Take $\Delta$ to be a singleton and we get the usual consequence relation whose range are single formulas.

Under this generalized notion of consequence Restall's remark about $\Gamma \models$ comes out trivially true. In the generalized setting $\Gamma \models$ iff $\Gamma \models \varnothing$ iff any model satisfying everything in $\Gamma$ satisfies something from $\varnothing$. Since $\varnothing$ is empty the last condition holds iff $\Gamma$ is unsatisfiable.

Answer 2

Expanding on my comment to make it a little more accurate and to elaborate on why it is the way it is, I stated: $\vDash\, : \mathcal{P}(\Omega)\times\mathcal{P}(\Omega)\to\mathbf{2}$ is a relation where I'm writing $\Omega$ for the set of formulas, $\mathcal{P}$ for the (finite) powerset operation, and $\mathbf{2}$ for the two-element set $\{\bot,\top\}$. I then wrote $X\vDash A\iff (\bigwedge\!X\to\bigvee\!A)$. This was more of an intuition than an actual definition. That said, it doesn't take much to make it into an actual definition. Equip $\mathbf{2}$ with its usual Boolean algebra structure. Assume a Boolean algebra homomorphism $[\![\_]\!] : \Omega\to\mathbf{2}$ which itself can be determined by a function $[\![\_]\!]_\mathbb A : \mathbb{A}\to\mathbf{2}$ where $\mathbb{A}$ is the set of atomic formulas. Then we can say $X\vDash A\iff (\bigwedge_{\varphi\in X}[\![\varphi]\!]\to\bigvee_{\psi\in A}[\![\psi]\!])$. In fact, we can generalize $\mathbf{2}$ to any Boolean algebra $\mathcal{B}$. At any rate, the function $[\![\_]\!]_\mathbb{A}$ is likely what Restall is calling an "evaluation".

But forget all that. A different, more syntactic approach would be to define a relation $\vdash_\mathbb A\, : \mathcal{P}(\mathbb{A})\times\mathcal{P}(\mathbb{A})\to\mathbf{2}$ as $X\vdash_\mathbb{A} A\iff X\cap A\neq\emptyset$. Now, for a pair of sets of arbitrary formulas, $X$ and $A$, we want to define $X\vdash A$ by reducing it to $\vdash_\mathbb{A}$. I use $\vdash$ as opposed to $\vDash$ as this is a logical/syntactic approach, not a semantic one. Of course, these two approaches are closely related. We have the following rules: $$ X\cup\{\bot\}\vdash A \iff \text{true} \iff X\vdash A\cup\{\top\} \\ X\cup\{\top\}\vdash A \iff X \vdash A \iff X\vdash A\cup\{\bot\} \\ X\cup\{\varphi\land\psi\}\vdash A \iff X\cup\{\varphi,\psi\}\vdash A \qquad X\vdash A\cup\{\varphi\lor\psi\} \iff X\vdash A\cup\{\varphi,\psi\} \\ X\cup\{\varphi\lor\psi\}\vdash A\iff (X\cup\{\varphi\}\vdash A\text{ and }X\cup\{\psi\}\vdash A) \\ X\vdash A\cup\{\varphi\land\psi\} \iff (X\vdash A\cup\{\varphi\}\text{ and }X\vdash A\cup\{\psi\}) \\ X\cup\{\neg\varphi\}\vdash A \iff X\vdash A\cup\{\varphi\} \qquad X\vdash A\cup\{\neg\varphi\} \iff X\cup\{\varphi\}\vdash A$$

Using the rules above, we see that $X \vdash A \iff \{\bigwedge\!X\}\vdash\{\bigvee\!A\}$. Let $\varphi\equiv\bigwedge X$ and $\psi\equiv\bigvee A$. We can put $\varphi$ into disjunctive normal form, and $\psi$ into conjunctive normal form. That is, $\varphi=\bigvee_m((\bigwedge_j p_{m,j})\land(\bigwedge_k \neg q_{m,k}))$, and $\psi=\bigwedge_n((\bigvee_j r_{n,j})\lor(\bigvee_k \neg s_{n,k}))$ where $p_{m,j}$, $q_{m,k}$, $r_{n,j}$, and $s_{n,k}$ are atomic formulas, i.e. elements of $\mathbb{A}$. (For the purpose of the normal forms, $\top = p\lor\neg p$ and $\bot=p\land\neg p$ for some $p\in\mathbb{A}$.) Applying the rules above, we ultimately get: $X\vdash A \iff \forall m,n.\{p_{m,j}\}_j\cup\{s_{n,k}\}_k\vdash_\mathbb{A}\{r_{n,j}\}_j\cup\{q_{m,k}\}_k$. The symmetry of the rules above is a major motivator for the definition of $\vDash/\vdash$.

Showing how these perspectives relate and expanding upon them is likely a large part of the content of Restall's Introduction, hence the "it will become clear" in my comment. The above is likely a decent chunk of the book being presented in a few paragraphs, so it probably isn't the most accessible thing at this point.

Answer 3

We can try to derive them starting from the "basic" definition of:

$v \vDash X$

where $X$ is a set of formulas and $v$ is an evaluation, meaning that the evaluation $v$ satisfies every formulas in $X$ (i.e. "for every formula $\varphi$, if $\varphi \in X$, then $v$ satisfy $\varphi$").

Thus, the def of $X \vDash A$ will be [page 31]:

for every $v$, if $v \vDash X$, then $v \vDash A$.

This can be rewritten as:

there is no evaluation $v$ such that: $v \vDash X$ and $v \nvDash A$.

Thus, with $X=\emptyset$, for every evaluation $v$: $v \vDash \emptyset$ [see page 40: "After all, there is nothing in the empty set for an evaluation to make false !" or written differently: "for every formula $\varphi$, if $\varphi \in \emptyset$, then $v$ satisfy $\varphi$" is vacuously true].

Thus, the condition: there is no evaluation $v$ such that $v \vDash \emptyset$ and $v \nvDash A$ will hold only when there is no evaluation $v$ such that $v \nvDash A$, i.e. for every evaluation: $v \vDash A$, i.e.:

$\emptyset \vDash A$ iff the formula $A$ is a tautology.

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In the same way, for $X \vDash \emptyset$, we have that:

there is no evaluation $v$ such that: $v \vDash X$ and $v \nvDash \emptyset$.

Now, for every valuation $v$: $v \vDash \emptyset$, is equivalent to: does not exist a valuation $v$ such that $v \nvDash \emptyset$.

Thus to say that there is no evaluation $v$ satisfying every formulas in $X$ means that $X$ is unsatisfiable, i.e:

$X \vDash \emptyset$ iff the set $X$ of formulas is a unsatisfiable.