I have to show the following

  1. Is Pr $[\varphi \leftrightarrow \psi]=1,$ so is $\operatorname{Pr}[\varphi]=\operatorname{Pr}[\psi]$

  2. Is $\operatorname{Pr}[\varphi \rightarrow \psi]=1,$ so is $\operatorname{Pr}[\varphi] \leq \operatorname{Pr}[\psi] $

  3. For all $\varepsilon>0$ $\operatorname{Pr}[\varphi \rightarrow \psi] \geq 1-\varepsilon,$ implies that $\operatorname{Pr}[\varphi] \leq \operatorname{Pr}[\psi]+\varepsilon .$

  4. In none of the cases holds the reversal . (For the third case Fall this means, that there is a distribution, formulas and an $\varepsilon>0$ exists with $\operatorname{Pr}[\varphi] \leqq \operatorname{Pr}[\psi]+\varepsilon$ but $\operatorname{Pr}[\varphi \rightarrow \psi]<1-\varepsilon $

Could someone please help me through that?


For the second one i did the following

$P[\varphi \rightarrow \psi]=1 \equiv P[\neg \varphi \vee \psi] = 1$

$ \equiv P[\neg \varphi] + P[\psi] - P[\neg \varphi \wedge \psi]=1$

$\equiv 1-P[\varphi] + P[\psi] - P[\neg \varphi \wedge \psi] = 1$

$\equiv P[\psi] - P[\neg \varphi \wedge \psi] = P[\varphi]$

$\implies P[\varphi] \leq P[\psi]$

Now i am stuck with the first case wherer i have $$P[\varphi \rightarrow \psi] \geq 1- \epsilon \equiv P[\neg \varphi \vee \psi] \geq 1 - \epsilon$$ $$\equiv P[\neg \varphi]+P[\psi]-P[\neg \varphi \wedge \psi] \geq 1- \epsilon $$ $$\equiv 1-P[\varphi]+P[\psi]-P[\neg \varphi \wedge \psi] \geq 1 - \epsilon $$ $$\equiv P[\psi]-P[\neg \varphi \wedge \psi] \geq - \epsilon + P[\varphi]$$ $$\equiv P[\psi] + \epsilon \geq P[\varphi] + P[\neg \varphi \wedge \psi]$$

  • $\begingroup$ Anyone some advice on 4. ? $\endgroup$ – Rapiz Jan 4 '20 at 19:06

Notice: If $p\geq q+r$ and $r\geq 0$, then $p\geq q$.

You have reached $\Pr[\psi]+\epsilon\geq\Pr[\varphi]+\Pr[\neg\varphi\wedge\psi]$, and so ...

  • $\begingroup$ Well yeah that simply states that Pr[ψ]+ϵ≥Pr[φ] thanks any ideas for the 4 ? $\endgroup$ – Rapiz Jan 1 '20 at 18:21
  • $\begingroup$ i have now done 1 to 3 i know that for the fourth i have to show that these do not hold backwards for each $\endgroup$ – Rapiz Jan 2 '20 at 19:56

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