Showing Consistency and Completeness implies a unique truth assignment

Question

I am struggling with the following problem. nb: this is all in propositional logic (P).

Show that the following are equivalent:

(1). $\Gamma $ is consistent and complete.

(2). There is exactly one truth assignment that satisfies $ \Gamma $.

My weak attempt:

$\Rightarrow$ Direction:

As $ \Gamma $ is consistent, we have that $ \not \exists A, \text{formula}, st ~\Gamma \vdash A ~\text{and}~\Gamma \vdash \neg A $

As $ \Gamma $ is complete, we have that $ \forall A, \text{formula}, \Gamma \vdash A ~\text{or}~ \Gamma \vdash \neg A $

truthfully,I struggle with moving past these definitions! Especially getting from he syntax to the semantics of it all, in order to incorporate truth assignments.

Please Help!

thank you:)

Answer 1

Recall that a truth assignment is a mapping that assigns a truth value (one of $f$ or $t$) to each propositional constant in your language. This is then extended to formulae recursively, via equations such as $v(A \wedge B) = v(A) \wedge v(B)$ and $v(\neg A) = \neg v(A)$.

You will need to recall and use the following facts:

1. $\Gamma$ is consistent precisely if there is some truth assignment $v$ such that satisfies $\Gamma$.
2. $\Gamma \vdash \psi$ precisely if $\Gamma \models \psi$. The latter, by definition, means that all truth assignments satisfying $\Gamma$ assign $v(\psi) = t$.

First, assume that $\Gamma$ is consistent and complete. Then, by the first fact above, there is some truth assignment $v$ that satisfies $\Gamma$. We will prove that there is only one. Take an arbitrary propositional constant $P$. By completeness, we have either $\Gamma \vdash P$ or $\Gamma \vdash \neg P$. By the second fact, $\Gamma \models P$ or $\Gamma \models \neg P$. In the first case, all assignments that satisfy $\Gamma$ assign $v(P) = t$, in the second case all assignments that satisfy $\Gamma$ assign $v(P) = f$. Since $P$ was arbitrary, all assignments assign the same truth value to all propositional constants, and hence all truth assignments are equal.

Now, assume that there is exactly one truth assignment $v$ satisfying $\Gamma$. By the first fact, $\Gamma$ is consistent. To show that $\Gamma$ is complete, take an arbitrary formula $A$. Evaluate $v(A)$. If $v(A) = t$, then in fact all truth assignments $w$ satisfying $\Gamma$ assign $w(A) = t$ (since there is only one such assignment, namely $v$). Thus, we have $\Gamma \models A$, and by the second fact, $\Gamma \vdash A$. Similarly, if $v(A) = f$, then $\Gamma \vdash \neg A$. Since $A$ was arbitrary, we get that $\Gamma$ is complete.