# Probabilities and propositional logic

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 $$\operatorname{Pr}[\varphi \rightarrow \psi]<1-\varepsilon$$

Edit1:

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 $$P[\varphi \rightarrow \psi] \geq 1- \epsilon \equiv P[\neg \varphi \vee \psi] \geq 1 - \epsilon$$ $$\equiv P[\neg \varphi]+P[\psi]-P[\neg \varphi \wedge \psi] \geq 1- \epsilon$$ $$\equiv 1-P[\varphi]+P[\psi]-P[\neg \varphi \wedge \psi] \geq 1 - \epsilon$$ $$\equiv P[\psi]-P[\neg \varphi \wedge \psi] \geq - \epsilon + P[\varphi]$$ $$\equiv P[\psi] + \epsilon \geq P[\varphi] + P[\neg \varphi \wedge \psi]$$

• Anyone some advice on 4. ? – 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 ...