MHB Trouble with understanding section of FOL completeness proof

  • Thread starter Thread starter pooj4
  • Start date Start date
  • Tags Tags
    Proof Section
Click For Summary
The Completeness Proof for First-Order Predicate Logic asserts that a consistent set of formulas, denoted as $\Phi$, is satisfiable. The process begins by transforming a consistent set $\Gamma$ into a maximally consistent set $\Gamma^*$ by adding either a formula or its negation while maintaining consistency. An interpretation $I$ is defined where the domain consists of all closed terms, and it satisfies a formula if that formula is part of $\Gamma^*. The proof then shows that $I$ satisfies all formulas $A$ if and only if $A$ is in $\Gamma^*$, ensuring that $I$ satisfies $\Gamma$. Understanding this proof may benefit from examining the induction on $A$ and the relevant properties of $\Gamma^*$ as outlined in Machover's text and "The Open Logic Text."
pooj4
Messages
4
Reaction score
0
The Completeness Proof for First-Order Predicate Logic depends on if $\Phi$ is a
set of consistent $\mathcal L$-formulas, then $\Phi$ is satisfiable.

How is that constructed? There are a large number of Lemmas working from Machover's text Set theory, Logic and Their Limitations but I'm having trouble with which are most relevant and how it comes together.
 
Physics news on Phys.org
The first step is to turn a consistent set $\Gamma$ into maximally consistent set $\Gamma^*$, i.e.., for each formula $A$ add either $A$ or $\neg A$ to $\Gamma$ in a way that preserves consistency. Then one defines an interpretation $I$ where the domain is the set of all closed terms and $I\models P(t_1,\ldots,t_n)$ if $P(t_1,\ldots,t_n)\in\Gamma^*$. Finally one proves that $I\models A\iff A\in\Gamma^*$ for all formulas $A$, not just atomic. Therefore $I\models\Gamma^*$ and in particular $I\models\Gamma$ since $\Gamma\subseteq\Gamma^*$.

I recommend starting with the last step. Try proving $I\models A\iff A\in\Gamma^*$ by induction on $A$ and see what this requires of $\Gamma^*$. Some steps go through for an arbitrary $\Gamma^*$; others require properties like completeness or existential completeness, which are ensures by the lemmas you mentioned. Also the book "The Open Logic Text" (Complete Build) has an outline of the proof in section 19.2.
 
First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K