Show $\vDash \phi \to \psi \Leftrightarrow \{\phi\} \vDash \psi$.

Question

Dirk van Dalen. "Logic and Structure (Universitext)" (p. 29)

Definition 1.2.1 A mapping $v : PROP \to \{0, 1\}$ is a valuation if
$
v(\phi \land \psi) = min(v(\phi), v(\psi)),\\
v(\phi \lor \psi) = max(v(\phi), v(\psi)),\\
v(\phi \to\psi)=0 \leftrightarrow v(\phi)=1 \text{and} v(\psi)=0,\\
v(\phi \leftrightarrow \psi)=1 \leftrightarrow v(\phi)=v(\psi), v(\lnot\phi) = 1 − v(\phi)\\
v(\bot) = 0\\
$

Definition 1.2.4
(i) $\phi$ is a tautology if $ [[\phi]]v$ = 1 for all valuations $v$,
(ii) $\vDash \phi$ stands for ‘$\phi$ is a tautology’,
(iii) Let $\Gamma$ be a set of propositions, then $\gamma \vDash \phi$ iff for all $v$ : $([[\phi]] v = 1 \text{for all } \psi \in \Gamma) \to [[\phi]]v = 1$.

My proof skeleton of one side of the proof: $\{\phi\} \vDash \psi \Rightarrow \, \vDash \phi \to \psi$.

Since $\{\phi\} \vDash \psi$, I know that for all valuations $v$, $[[\phi]]_v = 1 \Rightarrow [[\psi]]_v = 1$.
Proof:

• I start assuming $[[\phi]]_v = 1$
  • $[[\phi]]_v = 1 \to [[\psi]]_v = 1$
  • $[[\psi]]_v = 1$
• $[[\phi]]_v = 1 \to [[\psi]]_v = 1$
• $\vdots$

Am I on the right track ?

Answer 1

Starting with the assumption that $[[\phi]]_v = 1$ is correct.
The $[[\psi]]_v$ in your first sub-bullet point is strange; that's just an unknown number (a yet to be determined truth value) standing there, but after an "if ... then" one expects a statement. So just do without the sub bullet points and conclude $[[\psi]]_v = 1$ directly.
You should also generally add brief justifications how you obtain your results: Here, you used the assumption that $\phi \vDash \psi$.
Afterwards, you want to use this result to conclude that the implication $\phi \to \psi$ is true under the given valuation, justified by definition 1.2.1.

To complete the proof for the first direction, you then have to cover the other case: $[[\phi]]_v = 0$. That is, you do a proof by cases on the possible truth values of $\phi$, and obtain that the implication follows in eihter case.

Finally, you should make it clear what that $v$ is you are talking about: You are carrying out the proof for an arbitrary $v$, then conclude that since $v$ was arbitary, the above holds for all valuations, hence $\vDash$.

Taking this together, an improved version of your attempt looks as follows:

Assume $\phi \vDash \psi$.
Let $v$ be an arbitrary valuation.
There are two cases to distinguish:

1. $[[\phi]]_v = 1$.
   By the assumption $\phi \vDash \psi$, it follows that $[[\psi]]_v$ = 1.
   Then by the truth table of implication, $[[\phi \to \psi]]_v = 1$.
2. $[[\phi]]_v = 0$.
   $\vdots$

In both cases it holds that $[[\phi \to \psi]]_v = 1$.
Since $v$ was arbitrary, the above holds for all valuations, hence $\vDash \phi \to \psi$.