As you have proved, $\{\phi,\psi\}\vdash\psi\land\phi$ (since $\{\phi,\psi\}=\{\psi,\phi\}$ and according to point 3 of your exercise). And according to point 4 this means that $\{\phi\land\psi\}\vdash\psi\land\phi$. Using the deduction theorem we gain $\vdash(\phi\land\psi)\rightarrow(\psi\land\phi).$ Similarly, $\vdash(\psi\land\phi)\rightarrow(\phi\land\psi)$.
Now it's quite simple to conclude $\vdash(\phi\land\psi)\leftrightarrow(\psi\land\phi).$
I want to prove $\phi \models \psi$ iff $\phi \cup \{\neg \psi\}$ unsatisfiable.
What's the meaning of the union operator w.r.t. satisfiability in FOL?
We have to be careful with symbols. $a\models b$ could mean one of two things:
- $a$ is a model, and the formula $b$ is true for $a$.
- $a$ is a set of formulas, and for every model $\mathcal M$ that satisfies $a$, it must satisfy $b$.
In your case, it seems you mean $\phi$ is either a formula or a set of formulas. If $\phi$ is a set of (well-formed) formulas that are understood as axioms, then $\cup$ is literally the union of the two sets, that is to add $\neg\psi$ as a new axiom.
Now the statement is almost trivial: if $\phi\models \psi$, i.e. for any model $\mathcal M$ that satisfies $\phi$, it must satisfy $\psi$, hence it cannot satisfy $\neg\psi$, so $\phi\cup\{\neg\psi\}$ cannot be satisfied. If $\mathcal M$ satisfies $\phi\cup\{\neg\psi\}$, it satifies $\phi$ but not $\psi$, which contradicts the meaning of $\phi\models\psi$.
This is trivial precisely because $\vDash$ is about semantics. Only "$\phi\vdash\psi$ iff $\phi\cup\{\neg\psi\}$ cannot be satisfied" needs the Godel completeness theorem. See the difference between $\vdash$ and $\vDash$.
There seem to be many notational problems. Just to name a few, but you probably should read entire sections of your study materials more carefully.
So $\phi \cup \{\neg \psi\} \Leftrightarrow \phi \wedge \neg \psi$ ?
$\phi\wedge\neg\psi$ makese no sense, unless $\phi$ is a single formula instead of a (potentially infinite) set of formulas.
$\forall \varphi:(\beta\models \varphi) \wedge (\beta \models \neg\psi)$.
"$\forall, \wedge$" are reserved symbols in the FOL, so better to avoid using them on the meta-level. If $\beta$ is actually a model/interpretation (together with assignment for free variables if necessary), it's correct to use $\vDash$ here, but this $\vDash$ has a different meaning from the one in $\phi\vDash\psi$. Also, are you suggesting $\varphi\in\phi$?
If $\phi \cup \{ \neg \psi\} \models \phi$ can't be satisfied, then it has no models and so $\phi \cup \{ \neg \psi\} \models \bot$
It makes no sense to say whether "$a\vDash b$" can be satisfied, which is either true or false objectively. Only a set of sentences can be satisfied or not.
Best Answer
Rain and precipitation may be thought of as defined by unary predicates $R, P$ respectively true of precisely that which is rain and precipitation. To say that rain is not the same as precipitation essentially means $\neg \forall x (R(x) \leftrightarrow P(x))$. This is equivalent to $\exists x \neg (R(x) \leftrightarrow P(x))$, which by the result of the exercise implies $\exists x( \neg R(x) \leftrightarrow P(x))$. Glossing, if rain is not the same as precipitation, there's something which is not rain if and only if it is precipitation. (Snow, for instance, is such a thing.) That there are statements like $\exists x( \neg R(x) \leftrightarrow P(x))$ which don't gloss into very simple statements of ordinary language suggests the language of logical symbolism is sometimes more expressive than ordinary language.
To say that non-rain is the same as precipitation would be to say $\forall x( \neg R(x) \leftrightarrow P(x))$.