I am working on the following exercise from Dummit and Foote and I just want to make sure my logic is correct: Let $H,K$ be subgroups of $G$. Prove that $H \cup K$ is a subgroup iff $H\subseteq K$ or $K \subseteq H.$
I am attempting to prove that if $H\neq K$ and $H\subset K,$ then $K\subset H.$ By the logic given in the title, will this prove the forward implication of the exercise (I have already proven the converse)? I don't know anything about manipulating logical statements in the propositional calculus, so I wanted to make sure my intuition is correct.
With respect to the original question, we can first assume p. Then deduce the right side of the longest conjunction. Then apply the definition (x$\rightarrow$y) defined as ($\lnot$x $\lor$ y). So we have ($\lnot$($\lnot$a$\land$$\lnot$b$\land$p) $\lor$c). If we assume c, then ((a$\lor$b)$\lor$c). If we assume ($\lnot$($\lnot$a$\land$$\lnot$b$\land$p)), then (a$\lor$b$\lor$$\lnot$p). If we assume $\lnot$p, then we have a contradiction, so ((a$\lor$b)$\lor$c) follows. If we assume (a$\lor$b), then ((a$\lor$b)$\lor$c) follows. In either case, ((a$\lor$b)$\lor$c) follows. In either case, ((a$\lor$b)$\lor$c) follows.
So, yes, your intuition is correct.