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Remember that: $(a\implies b) \equiv (\neg a \lor b)$ Hence: $\therefore \neg (a\implies b) \equiv (a \land \neg b)$ The counter example of the implication is an event where $a$ happens and not $b$. ${\begin{array}{|c|c|c|c|}\hline a & b & a\to b & a \land \neg b \\ \hline\hline F & F & T & F \\\hline F & T & T & F \\ ... 2 You can disprove the statement$a \implies \sim b$using a counterexample if you can find one. However, this does not prove the statement$a \implies b$. This is because the negation of$a \implies \sim b$isn't$a \implies b$. Recall that the negation of an implication is not an implication.$\sim (a \implies b) \equiv a \wedge \sim b$. In words, this ... 3 Instead of having both$A$,$B$,$a$and$b$, let's make some of them$p$and$q$. Also, since you're talking about counterexamples, there must be a quantifier somewhere, so I guess you actually mean something like $$A \equiv \forall x\,(p(x)\to q(x))$$ $$B \equiv \forall x\,(p(x)\to \neg q(x))$$ Having a counterexample to$A$means that we have a ... 1 If$|a|>1$then $$\quad0<\frac{1}{|a|}\text{ and } \frac{1}{|a|}<1.$$ So,$\frac{1}{|a|}$is a number between$0$and$1$. Since there is no integer between$0$and$1$, we conclude that$\frac{1}{|a|}\notin\mathbb Z$. EDIT: Proof by contradiction, as you want: Suppose$\frac{1}{|a|}$is an integer. Since$|a|>1$, we conclude that ... 5 Suppose$\frac{1}{a} = b$an integer. I multiply both sides of this equation by a, giving$ab = 1.$This implies$a = b = \pm 1$since the only integers who multiply out to$1$are$\pm 1$(since they are the units in the ring of integers) . But$|a|>1\$. This is a contradiction.