# Confusion on Section 1.2 of Rosen's Discrete Math Textbook

So I was able to deduce based on the rule that p implies q is the same as q unless (not p) that this is same as:

(not s) -> (r -> (not q))

I could use the logical equivalence (A -> B) = (A or (not B)) to get Rosen's result, but up to this part of the textbook he didn't introduce logical equivalences like this one.

So I'm just confused if there was a different way he wanted us to get to this answer. Or is it that he just is telling us the answer and not asking us to derive it ourselves but instead just believe it?

• Note $A\rightarrow B$ is equivalent to $(\lnot A)\lor B$, not to $A\lor(\lnot B)$ as you wrote. Commented May 17 at 0:51
• In a recent thread here, it was deduced that "q unless not p" is equivalent to "p if and only if q." Commented May 17 at 13:47

As you probably know, you can derive the formula $$( r \wedge \neg s ) \to \neg q$$ from $$\neg s \to (r \to \neg q)$$ by means of equivalency laws
$$\begin{array}{llll} & \neg s \to (r \to \neg q) & \text{Premise} \\ \equiv & \neg \neg s \vee (r \to \neg q) & \text{Implication Rule} \\ \equiv & \neg \neg s \vee (\neg r \vee \neg q) & \text{Implication Rule} \\ \equiv & (\neg \neg s \vee \neg r) \vee \neg q & \text{Associativity of \vee} \\ \equiv & \neg ( \neg s \wedge r) \vee \neg q & \text{DeMorgan's Law} \\ \equiv & ( \neg s \wedge r) \to \neg q & \text{Implication Rule} \\ \equiv & ( r \wedge \neg s ) \to \neg q & \text{Commutativity of \wedge} \\ \end{array}$$
$$\begin{array}{llll} \{1\} & 1. & \neg s \to (r \to \neg q) & \text{Premise} \\ \{2\} & 2. & r \wedge \neg s & \text{Assumption for Conditional Proof} \\ \{2\} & 3. & \neg s & \text{2 \wedge Elimination} \\ \{1,2\} & 4. & r \to \neg q & \text{1,3 Modus Ponens} \\ \{2\} & 5. & r & \text{2 \wedge Elimination} \\ \{1,2\} & 6. & \neg q & \text{4,5 Modus Ponens} \\ \{1\} & 7. & (r \wedge \neg s) \to \neg q & \text{2,6 Conditional Proof} \\ \end{array}$$