0

My question is whether $A \Rightarrow \forall x(A(x))$ is provable in the classical predicate calculus. My intuition would say yes because using finite domain truth tables one can check (I hope I am not wrong) that $A \Rightarrow \forall x(A(x))$ is a tautology. Because if $A$ evaluates false for some variable then $\forall x(A(x))$ is false so implication is true and if $A$ evaluates true for all variables then both $A$ for each variable is true and so is $\forall x(A(x))$, so implication is true as well.

I am confused because in Kleene's "Mathematical Logic" he says that $A \vdash \forall x (A(x))$ is not true. Here, he means that $A \vdash B$ if $B$ is deducible from $A$ when all free variables of $A$ are held constant. I think this statement contradicts the fact that $A \Rightarrow \forall x(A(x))$ is provable due to the following reasoning.

  1. $A \Rightarrow B, A \vdash B$ is a valid deduction in predicate calculus.
  2. Then, $A \Rightarrow \forall x(A(x)), A \vdash \forall x(A(x))$ is a valid deduction.
  3. Because $A \Rightarrow \forall x(A(x))$ is provable, it means $A \vdash \forall x(A(x))$.

I would appreciate if you can help me with this.

Daniels Krimans
  • 1,473
  • 1
    You mistake is to assume that $A \to \forall x (A(x)$ is provable in Kleene's system. See page 107 : "$A \vDash \forall x A$ does not hold in general". – Mauro ALLEGRANZA Mar 31 '19 at 08:26
  • 1
    In Kleene's system we have instead (same page) $A \vDash^x \forall x A$ and we can apply the Deduction Th (page 112) only if "all variables are held constant". – Mauro ALLEGRANZA Mar 31 '19 at 08:28
  • 1
    In a nutshell, $A \to \forall x A(x)$ is not provable(in every sound calculus) because it is not valid. See Kleene counterexample page 102. The most "natural" example is : $(x=0) \nvDash \forall x (x=0)$. – Mauro ALLEGRANZA Mar 31 '19 at 08:34

1 Answers1

3

What $P\vdash Q$ should mean when $P$ and $Q$ have free variables is handled in different ways for different presentations of classical (and other) logics. For Hilbert-style presentations of logic, it is often defined to be $\forall P\vdash\forall Q$ where $\forall P$ means the universal closure of $P$, i.e. $P$ with all free variables bound by universal quantifiers.

In other presentations it makes less sense to do this, and even for Hilbert-style presentations you can make a different choice which Kleene does. One hint on what it should mean from Kleene's definition of $\vdash$ is in the "held constant". One way of viewing this is that when considering $P\vdash Q$ where free variables occur in $P$ and/or $Q$ is that those free variables become constants of your signature. That is, $P(x)\vdash Q(x)$ means, "add $\mathsf{c}$ to the signature as a constant and see if $P(\mathsf c)\vdash Q(\mathsf c)$, which now only talks about closed formulas, is derivable.

For your specific case, it should then be clear why $A(y)\vdash\forall x.A(x)$ isn't derivable. $A(1)$, say, doesn't imply $\forall x.A(x)$. Your "proof" fails because, unsurprisingly $A\Rightarrow\forall x.A(x)$ isn't a "tautology". Your semantic argument fails because if $A$ holds for some but not all elements in the domain, then there is an interpretation of $A(\mathsf c)\Rightarrow \forall x.A(x)$ that interprets $\mathsf c$ as a value for which $A$ holds but $\forall x.A(x)$ producing "true" implying "false". Again, you wouldn't expect $A(1)\Rightarrow \forall x.A(x)$ to be valid.

Derek Elkins left SE
  • 24,967