How to show that $p \oplus p$ is a contradiction?

Question

One of my textbook solutions shows why $p \oplus p$ is a contradiction with something like these steps:

$$\begin{align*}p \oplus p &\quad& {}\\ (p \vee p) \wedge \lnot(p \wedge p) & & \text{by definition} \\ p \wedge \lnot p & & \text{ by the identity laws}\end{align*}$$

The above is clearly a contradiction.

I don’t understand how they get from the second line to the third. From what I understand, the identity law is that $p \vee T$ is equivalent to $p,$ but I don’t see how this gets us from line 2 to line 3.

Answer 1

Formally, if you aren't able to say $p \wedge p = p,$ then maybe you know that $\wedge$ and $\vee$ distribute over each other? In that case, you could see $$p \wedge p \equiv (p \vee F) \wedge (p \vee F) \equiv p \vee (F \wedge F) \equiv p \vee F \equiv p$$

Answer 2

It is from the idempotence identities, which are $p\vee p\equiv p$ and $p\wedge p\equiv p$.

So $\quad(p\vee p)\wedge\neg(p\wedge p)\\ \equiv (p)\wedge\neg(p)$

That is all.

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You may be encountering a confusion of terminology.   Truth ($\top$) is known as the conjunctive identity, and falsity ($\bot$) as the distributive identity .   However, the identity laws are a list of fourteen equivalences used for substitutions in propositional calculus (including these).

Also, these are not to be confused with the first order logic's Law of Identity: $\forall x~(x=x)$.