0
$\begingroup$

enter image description here

This is a proof from the book Lectures in Logic and Set Theory: Vol 1. But I cannot understand why this proof is valid.

This is a corollary of "all first order logic is sound" which I understand the proof.

There is another proof of the consistency of first order logic in another book An Introduction to Mathematical Logic and Type Theory: To Truth Through Proof... but it is too long and I forget most of the book so I lost the context...

1 Answers 1

1

See page 56 : Definition I.5.9 (Soundness). A theory $\Gamma$ is sound iff, for all $\mathcal A ∈ \text {Wff}$ $\Gamma \vdash \mathcal A$ implies $\Gamma \vDash \mathcal A$.

To say that a first order theory $\Gamma$ is sound (page 58) means that the theory $\Gamma$ does not prove false formulae.

Thus (by contraposition): if $\Gamma \nvDash \varphi$, then $\Gamma \nvdash \varphi$.

But $\nvDash \lnot x=x$: the formula $x \ne x$ is clearly not valid. And thus, by soundness, the calculus cannot prove it, i.e. $\nvdash \lnot x=x$ (because every theorem of the calculus is also a provable in a theory $\Gamma$ whatever).

  • 0
    mmm... ok. I understand why I don't understand the proof. because the proof is a non-constructive one. this also easily proofs that a pure theory always has a model, but not making it explicit... – 2017-02-23
  • 0
    I was under the impression that all proofs done in meta-mathematics needs to be constructive.. – 2017-02-23
  • 0
    @MinghaoLiu - not sure to understand... The *calculus* (the "pure" logic) is consistent. Of course, it has models because the calculus proves only *valid* formulae (by soundness) and the valid formuale are (by definition) true in every model. Consider $x=x$; it is valid and thus it is true in every model. This does not necessarily means that a consistent theory, where a theory is the calculus + a set $\Gamma$ (finite or infinite) of axioms: the axioms specific to the theory, has a model. – 2017-02-23
  • 0
    In a nutshell, to add specific axioms to the calculus means to restrict the number of possible models. Consider e.g. Peano's axioms for arithmetic: they are not *valid* (true in every model): a structure (if any) that satisfy them is a *model* of the arithmetical theory. – 2017-02-23
  • 0
    I wanted to construct a explicit model of the pure logic, then because I can show that $x != x$ is false in this model, then I can proof that this formula is not universally valid. then proofing the pure logic is consistent. So maybe in the metamathematics, this model is trivial? – 2017-02-23
  • 0
    again, in what extent does this depends on the metamethematics? if the metamathematics is a constructive logic like Martion-Lof Type Theory, then this proof is not valid, because it uses contraposition? – 2017-02-23
  • 0
    @MinghaoLiu - the "usual" meta-theory of classical logic is non-constructive: it uses model, defined with set theory. There are constructive proof of consistency of FOL; see e.g. [Hilbert and Ackermann](https://books.google.it/books?id=45ZGMjV9vfcC&printsec=frontcover). – 2017-02-23
  • 1
    @MinghaoLiu - "explicit model of the pure logic" ? Every structure, also a finite one, is a "model" of pure logic, because the theorems of pure logic are true in **every** model. Thus the structure $\{ myself, my dog \}$ is a model of pure logic. Obviously I'm equal to myself (how we can have that I'm different from myself ?) and the same for my dog. Thus, in this structure : $x \ne x$ is false. But to prove that $x=x$ is valid it is not enough to show that it is *false* in some structure; we have to prove that it is *unsatisfiable* (i.e. *false* in every structure). 1/2 – 2017-02-23
  • 0
    Thus, if $\varphi$ is unsat, then $\lnot \varphi$ is *valid*. – 2017-02-23
  • 0
    I think we don't need to proof $x = x$ is valid, but only to proof $x != x$ is not universally valid, since "universally valid" is defined as "every structure for the language", so to proof that this is not universally valid, we only need to find one structure that the formula is not true? – 2017-02-23