Originally Posted by

**novice** The above should be ~(~p=>p). It's the same as saying, "It is false that {If 2+2 $\displaystyle \not=$4, then 2+2=4.}"

If p is true, then p v p is true. In logical expression $\displaystyle p \Rightarrow p \vee p$

The conjunction is true if at least one is true.

If the p v p is true, then p is true. In logical expression $\displaystyle p \vee p \Rightarrow p$

$\displaystyle (p \Rightarrow p \vee p) \vee (p \vee p \Rightarrow p)$ is equivalent to $\displaystyle p \Leftrightarrow p \vee p$.

Now $\displaystyle p \vee p \Leftrightarrow \sim \sim p \vee \sim \sim p$.

De Morgan's Law says $\displaystyle \sim \sim p \vee \sim \sim p\Leftrightarrow \sim (\sim p \wedge \sim p)$

$\displaystyle \sim (\sim p \wedge \sim p) \Leftrightarrow \sim (\sim p \Rightarrow p)$

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The conjunction $\displaystyle p \vee p$ is like you are being offered to have a potato or a potato. There is only one choice, potato.

It's different if the conjuction is $\displaystyle p \vee q$, which is like being offered rice or potato. If it's rice then it's not potato. So either way, you get at least one.

$\displaystyle p \Rightarrow p \vee p$ is like the same as saying, "If it's a potato, then you get potato or potato."

Here is the De Morgan's Law $\displaystyle \sim(P \vee Q) \equiv \sim P \wedge Q$.

Here is the negation of implication: $\displaystyle \sim (P \Rightarrow Q) \equiv P \wedge \sim Q$