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For arbitrary formulas $A,B,C$ it holds that:

  1. $\{A,B\} \vDash C $ if $A \vDash (B \Rightarrow C)$
  2. $(A \Rightarrow B) \vDash C$ if $A \vDash (B \Rightarrow C)$
  3. $A \vDash C$ if $A \vDash (B \Rightarrow C)$

I know that only first one holds, can someone explain me why?

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Here's one approach:

  1. Note that trivially $\vDash p \to p$, so a fortiori $p \vDash p \to p$. But $p \to p \nvDash p$ (suppose $p$ is false). So we can have an instance of $A \vDash B \to C$ without the corresponding $A \to B \vDash C$.

  2. Note that trivially $\vDash q \to q$, so a fortiori $p \vDash q \to q$. But of course $p \nvDash q$. So we can have an instance of $A \vDash (B \Rightarrow C)$ without the corresponding $A \vDash C$.

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  • $\begingroup$ You're welcome! $\endgroup$ – Peter Smith Apr 8 '13 at 19:13
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The canonical way to show that the two last implications don't hold would be to find formulas you can plug in for $A$, $B$ and $C$, such that the entailment to the right of the "if" is logically valid, but the one to the left isn't.

For example, try setting $B\equiv P$, $C\equiv Q$ and $A\equiv (P\Rightarrow Q)$, where $P$ and $Q$ are propositional variables.

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Number 3 Should be fairly obvious:

A entails (B implies C) 

Which, in English, becomes:

A entails C or A entails not B

Number 2 is a bit more complex. The values it fails for are $A, ¬B, ¬C$.

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