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I know that $⊨$ symbol is entails symbol and $A⊨B$ means that if A is True then B must be True.

But I'm confused about the $⊭$ symbol. which one is it?

  • $A⊭B$ means if A is true then B is False? = $A⊨¬B$
  • $A⊭B$ means the Trueness of A is not any guarantee for B?
  • $A⊭B$ means if A is False then B must be True? = $¬A⊨B$

thank you in advance.

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    $\begingroup$ Clearly not the third, as that is consistent with $A⊨B$ $\endgroup$ – Henry Jun 13 at 21:05
  • $\begingroup$ @Henry, one and three are consistent with that. the problem was I couldn't find any resource for it on the web! all is just about $⊨$. $\endgroup$ – Peyman mohseni kiasari Jun 13 at 21:09
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    $\begingroup$ @Peymanmohsenikiasari No, option 1 is not consistent with $A \vDash B$. If $A \vDash B$, then all assignments which make $A$ true must also make $B$ true, but then $\neg B$ can never be the case. $\endgroup$ – lemontree Jun 13 at 21:11
  • $\begingroup$ @Henry Oh! my mistake. thank you. $\endgroup$ – Peyman mohseni kiasari Jun 13 at 21:12
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$A \vDash B$ means

For all assignments $v$, if $A$ is true under $v$, then $B$ is true under $v$.

$A \not \vDash B$ simply is the (meta-logical) negation of this statement, that is

Not for all assignments $v$ it is the case that if $A$ is true under $v$, then $B$ is true under $v$

which is equivalent to

There is at least one assignment $v$ such that $A$ is true under $v$ but $B$ is not

which means your second option is the right one.

The other two options (1 and 3) are notated in the way you already figured out by yourself.

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  • $\begingroup$ well, so if $A⊭B$ then A must be satisfiable I think. nice, thank you. $\endgroup$ – Peyman mohseni kiasari Jun 13 at 21:11
  • $\begingroup$ @Peymanmohsenikiasari Correct. Because if $A$ is not satisfiable (= is a contradiction), then it entails any formula. $\endgroup$ – lemontree Jun 13 at 21:12
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    $\begingroup$ To emphasize, $A\nvDash B$ literally means "it is not the case that $A\vDash B$". That is, it is simply the meta-logical negation of $A\vDash B$. That is why there is not special discussion of it. We can, as lemontree does, work out a more pleasant, equivalent formulation. $\endgroup$ – Derek Elkins Jun 13 at 22:16

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