Let $a,b\in\Bbb{R}$. Show if $a\leq b_1$ for every $b_1>b$, then $a\leq b$. Solve using proof-by-contradiction and write out using logical premises.
Proof: Let $Q: a\leq b_1$ for every $b_1>b$ and $R:a\leq b$ then $P:Q\to R$. Suppose $\neg P$, then we have $\neg P\iff Q\land\neg R$. Thus suppose $a>b$ and let $b_1=\dfrac{a+b}{2}$ and consider $Q$. We have $b_1>b$ but $a>b_1$ hence $\neg R\to\neg Q$. Hence we have $$(\neg P\Rightarrow Q\land\neg R) \land(\neg P \Rightarrow\neg R\to\neg Q)\iff\neg P\Rightarrow F$$ and thus $P$ must be true. $\square$
I am uncertain about the last step where I have $\neg P \Rightarrow\neg R\to\neg Q$ and use it to combine the contradiction $(Q\land\neg R)\land(\neg R\to\neg Q)$.