0

$ψ \vDash ⊥$ 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?

Constantly confused
  • 1,548
  • https://math.stackexchange.com/questions/2431003/wondering-why-proof-by-contradiction-works?rq=1 The post in this link may not be fully right for you, but there are other links that can be found in it. – Mr Pie Mar 23 '18 at 00:28
  • I understand that there are many proof by contradiction questions around; but they seem to have assumed this theorem to be true, instead of trying to prove it...? – Constantly confused Mar 23 '18 at 00:29
  • Proof by contradiction is a technique to prove theorems. Could the link below be what you are looking for? $\longrightarrow$ https://math.stackexchange.com/questions/975470/show-that-gamma-cup-neg-phi-is-satisfiable-if-and-only-if-gamma-not?rq=1 – Mr Pie Mar 23 '18 at 00:46
  • Thank you for your help; but I don‘t think it‘s quite what I am looking for. I actually feel rather comfortable with solving that question – Constantly confused Mar 23 '18 at 01:01
  • 1
    Ok. I swear, though, there was another post that I went to which I believe would have been quite good for you, but I cannot find it. – Mr Pie Mar 23 '18 at 01:03
  • What definition are you using for "$\psi$ is a contradiction"? – Derek Elkins left SE Mar 23 '18 at 01:44
  • 1
    "so how can any truth assignment satisfy $\bot$ ?" Exactly: if we have $\psi \vDash \bot$, every truth assignment $v$ that satisfies $\psi$ must also satisfies $\bot$. But it can't. Concluson: if we have $\psi \vDash \bot$, necessarily $\psi$ must be always false i.e. unsatisfiable (like e.g $p \land \lnot p$). – Mauro ALLEGRANZA Mar 23 '18 at 07:18
  • @MauroALLEGRANZA Is this kind of like the equivalent of vacuous truth in conditionals...? The two sound very similar but I have never heard people talk about vacuous truth in entailment – Constantly confused Mar 25 '18 at 08:31
  • @DerekElkins The definition of contradiction as per the textbook I am using is that it is false under every truth assignment, would a different def. make any difference? – Constantly confused Mar 25 '18 at 08:37
  • 1
    Not exactly: vacuous truth may refers to a conditional statement with a false antecedent. Thus, it is linked to: $\bot \vDash \psi$, for every $\psi$. – Mauro ALLEGRANZA Mar 25 '18 at 08:56

1 Answers1

3

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" .

bof
  • 82,298
Graham Kemp
  • 133,231
  • Thank you! But the way you put it make it sounds like a vacuous truth...I don't suppose they are 'related'...? I mean, both are just as counter-intuitive in the same sense – Constantly confused Mar 25 '18 at 08:44
  • 1
    Yes, they are somewhat related. Basically, $\psi\models \bot$ means the set of assignments which satisfy $\psi$ must be a subset of the set of assignments which satisfies $\bot$, which is an empty set. – Graham Kemp Mar 25 '18 at 11:50