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I am self-studying logic, and am exploring the discrepancy that several texts (philosophical and mathematical) assert that quantifiers can't be assessed "semantically" (via a truth table), and a text by Ackermann that asserts that they can.

In Ackermann's 1954 Solvable Cases of the Decision Problem, on page 6 he writes the following:

V6. A wff $(\mathcal U)\mathfrak U ( \mathcal U)$ has the value T by an assignment for the free variables, if $\mathfrak U ( \mathcal U)$ has the value T by the same assignment from the free variables different from $ \mathcal U$and an arbitrary assignment for the variable $\mathcal U $; a wff $(\exists\mathcal U)\mathfrak U ( \mathcal U)$ has the value T for an assignment, if there is an assignment for $\mathcal U $ which together with the given assignment for the free variables of $(\exists\mathcal U)\mathfrak U ( \mathcal U)$ gives the value T to $\mathfrak U ( \mathcal U)$. In every other case, $(\mathcal U)\mathfrak U ( \mathcal U)$ and $(\exists\mathcal U)\mathfrak U ( \mathcal U)$ have the value F by an assignment.

I understand the part regarding the existential quantifier; however, it seems that for the universal he's claiming the bound variable(s) have no relevance?! e.g. $(x)(Rx \land y)$ can't be F if y is T?

I suspect my reading of Ackermann is wrong based on other texts I've read but, putting those aside, I'm not sure how else to read it. Indeed in a two-part example on pg.6-7 he demonstrates that this is the case. Yet in my counter example it is entirely possible for (x)(Rx $\land$ y) to be false... so I'm grossly confused, and find it demotivating to continue reading this book (in the second half of the example he uses a domain of just 2 members, a & b, and yet goes on to mention members c & d...sigh).

Lugh
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1 Answers1

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It says:

A wff $(x)A(x)$ has the value T by an assignment for the free variables, if $A(x)$ has the value T by the same assignment from the free variables different from $x$ and an arbitrary assignment for the variable $x$.

It means: consider a variable assignment $s : \text {Var} \to D$, where $D$ is the domain of the interpretation.

And consider a new assignment $s'$ such that $s'(y)=s(y)$ if $y$ is free in $A(x)$ and $y \ne x$ and $s'(x)$ whatever.

If $A(x)$ has value T for every such $s'$, then $(x)A(x)$ has value T for $s$.

Intuitively, $(x)A(x)$ is true if $A(x)$ is true for every value that we can assign to $x$.

Regarding your example:

$(x)(Rx∧y)$ can't be F if $y$ is T ?

we have to consider that $y$ is a variable; thus, an assignment $s$ assigns to $y$ an "object" of the domain, and not a truth value.

About:

Is he asserting that all wff with no (free) variables evaluate to F ?

Of course no. A sentence (i.e. a wff with no free variables) is evaluated to T (F) by an assignment $s$ iff every assignment evaluates it to T (F).

Consider e.g. the formula $(x)(x=0)$ and interpret it in the domain $\mathbb N$ of natural numbers.

Consider $s(x)=0$; by the above definition, it is not true that $s$ satisfies the formula in $\mathbb N$, i.e. $\mathbb N, s \nvDash (x)(x=0)$.

This is so becuase if we consider e.g. $s'$ such that $s'(x)=1$, we have that $(x)(x=0)$ is not satisfied by $s'$ (we have that: $0 \ne 1$) and thus it is not true that every assignment that differs from $s$ only on the value assigned to $x$ satisfies the formula.

Mauro ALLEGRANZA
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  • In reverse order: Sorry, the question about bound-wff evaluating to F was based on an example where he says "...has simply the value F, since no free variables occur." However reading further, where the same wff evaluated to T on a different domain, I realized my mistake. I attempted to edit my question by removing the paragraph where I point the above out, but apparently left the question itself behind. I have now removed the question, but I would that part of your answer to remain because it adds further clarity to the remaining question. – Lugh Jul 27 '17 at 14:03
  • I had wondered if my counter-example should have been something like (x)(Rx∧(y=x)). I appreciate your response, because it hammers home the point that first/second order logic is not restricted to truth values. However, (x)(Rx∧P), where P is a propositional variable, would stay inside {T, F}, yes?
    • – Lugh Jul 27 '17 at 14:10
  • I realize that I have seen before (and understood) similar definitions to the one provided by Ackermann, but the example he provided seems to have confused me. He discussed the sentence $(z)(\exists u)(\exists v)(z=u \lor z=v \land \overline{u=v})$ for 1 and 2 member domains ({a} and {a, b}). Am I correct in understanding that this sentence is false for any domain larger than 2? If so, then I see where I went wrong in reading Ackermann ("gloss over" is too harsh of a word, but some of his explanations have that feel, compared to say how you explain things).
    • – Lugh Jul 27 '17 at 14:30
  • @Lugh - for 1): wrong. Consider a domain ${ a,b,c }$: we have that $(a=a \lor a=b \land \lnot(a=b))$ is true. Thus, the formula is satisfied. – Mauro ALLEGRANZA Aug 01 '17 at 08:33
  • In order to impose tha cardinality on the doamin, we must have: $(\exists u)(\exists v) (\lnot (u=v) \land (\forall z)(z=u \lor z=v))$ – Mauro ALLEGRANZA Aug 01 '17 at 08:34