I have been trying to break down these two formula correctly using natural deduction, and now I am stuck and confused. Below there is my attempt to derive the propositional logic consequences. I need your help. Listing form is also acceptable. Thanks already.
\begin{align}
(1.) &\lnot p \to (p \land r) \vDash \lnot p \to r\\
(2.) &\vDash (p \to (q \to (p \to (p \lor q))))
\end{align}
Unsure Breakdown of what I did in question (1). Question (2) is hard for me:
I derived $\lnot p \to r$ with V-E, since $\lnot p$ is $\lnot p$, and $r$ can be derived from $p \land r$.
Best Answer
Since it is not clear which kind of formalism you use for natural deduction, I use the one I prefer, the tree-like one.
The idea of the proof is that, given the hypothesis $\lnot p \to (p \land r)$, if you suppose $\lnot p$ then you can conclude $r$, because by modus ponens (the rule $\to_E$) from the hypothesis $\lnot p \to (p \land r)$ and the further assumption $\lnot p$ you get $p \land r$ and in particular $p$ (by $\land_E$). So, if you discharge the further assumption $\lnot p$ (using the rule $\to_I$), you get a proof of $\lnot p \to r$ under the hypothesis $\lnot p \to (p \land r)$.
Formally, a derivation in natural deduction with hypothesis $\lnot p \to (p \land r)$ and conclusion $\lnot p \to r$ is the following:
\begin{align} \dfrac{\lnot p \to (p \land r) \qquad [\lnot p]^*}{\dfrac{\dfrac{p \land r}{r}\land_E}{\lnot p \to r} \to_I^*}\to_E \end{align}
Note that, since in the hypotheses and conclusion there are no occurrences of the connective $\lor$ (disjunciton), you need not use inference rules $\lor_E$ or $\lor_I$ here.