Simplifying Boolean expressions using De Morgan's laws and tautologies

Believing that $q\to (p\wedge q)$ is a tautology, confusing a conditional relation with an equivalence. Evaluate the truth value: when $q$ is true and $p$ is false, $q\to (p\wedge q)$ evaluates to $\text{True}\to \text{False}$, which is false, so it is not a tautology.

Assuming that $(\sim p\vee \sim q)\vee q$ reduces to $\sim p\vee q$, omitting the tautological nature of $\sim q\vee q$. Recognize that $(\sim q\vee q)$ is always true ($T$), making the entire disjunction $\sim p\vee T\equiv T$, whereas $p\to q$ can be false when $p$ is true and $q$ is false.

Confusing the nested implication $p\to (p\to q)$ with a tautology. Rewrite $p\to (p\to q)$ as $\sim p\vee (\sim p\vee q)\equiv \sim p\vee q$, which is not identically true.