0

Given the open formula: $\alpha =(\exists{{x}_{2}})({P}^{1}({x}_{1},{x}_{2}))$

And consider the interpretation $I$ where the domain is the natural numbers, and ${P}^{1}$ means equality.

Is $\alpha$ true, false, or indeterminate under $I$?

I think is true, because in any possible valuation of the free variable ${x}_{1}$, there exists ${x}_{2}$ such that ${x}_{1}={x}_{2}$. However my "official solution" says it is indeterminate.

Is my reasoning correct?.

Wyvern666
  • 911

1 Answers1

2

The question is "tricky"...

According to some textbook, an open formula (a formula with free variables) is true (in a structure) iff its universal closure is; see e.g. Dirk van Dalen, Logic and Structure (5th ed - 2013), page 67.

According to this definition, you are right: $\mathbb N \vDash \alpha$. In general, with $P^1$ interpreted as equality, the formula is true in every structure.

But there are other approaches; see e.g. Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001), page 84.

According to this approach, the interpretation of a formula $\varphi$ in a structure $\mathfrak A$ is defined with a variable assignment function $s : Var \to |\mathfrak A|$, where $|\mathfrak A|$ is the domain of $\mathfrak A$.

Thus, the basic semantic relation is:

$\mathfrak A, s \vDash \varphi$.

According to this definition, $\mathbb N \vDash \alpha$ is not defined, because we have not specified $s$.

Conclusion: you have to check the details of the semantic specifications of your textbook.

Mauro ALLEGRANZA
  • 94,169
  • The exercise have no valuation function defined. However, my class notes says that for open formulas: $I \vDash \varphi$ iif $I, v \vDash \varphi$ for every valuation $v$ under $I$. I thought that was somewhat universal, im sorry if it's not. So, im considering that for any valuation, the formula is true under given $I$. This makes sense? – Wyvern666 Jun 24 '16 at 18:20
  • 1
    @Wyvern666 - yes; it is the same as considering the universal closure. – Mauro ALLEGRANZA Jun 24 '16 at 20:27