$ψ \vDash ⊥$ if and only if $ψ$ is a contradiction - why?

Question

$ψ \vDash ⊥$ iff $ψ$ is a contradiction

I am confused as to why this is; the definition of $ψ \vDash \phi$ as I understand is, for all truth assignments that satisfies $ψ$, they also satisfy $\phi$. But $⊥$ will always be false - so how can any truth assignment satisfy it? It seems to make no sense to say anything can entail $⊥$.

Intuitively for $\leftarrow$, I get that if $\psi$ is a contradiction then its truth value will always equal that of $⊥$ - but then that's still not quite the same as an entailment.

As for $\to$,as I mentioned above the notion of $ψ \vDash ⊥$ just doesn't make sense to me.

Could anyone please help?

Answer 1

I am confused as to why this is; the definition of $ψ⊨ϕ$ as I understand is, for all truth assignments that satisfies $ψ$, they also satisfy $ϕ$. But $⊥$ will always be false - so how can any truth assignment satisfy it?

Exactly.   That is point of the argument.

• Since $\bot$ can never be satisfied, therefore there must never be assignments which will satisfy $\psi$ if we can truly say "every assignment which satisfies $\psi$ also satisfies $\bot$".

  That is: $\psi$ is a contradiction if $\psi\vDash\bot$.

• Since "$\psi$ is a contradiction" means $\psi$ can never be satisfied, therefore we can truly say "every assignment which satisfies $\psi$ also satisfies $\bot$" if $\psi$ is a contradiction.

  That is: $\psi\vDash\bot$ if $\psi$ is a contradiction.

• Thus it is "if and only if" .