Rosen 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?

  • 1
    $\begingroup$ Note $A\rightarrow B$ is equivalent to $(\lnot A)\lor B$, not to $A\lor(\lnot B)$ as you wrote. $\endgroup$
    – blargoner
    Commented May 17 at 0:51
  • $\begingroup$ In a recent thread here, it was deduced that "q unless not p" is equivalent to "p if and only if q." $\endgroup$ Commented May 17 at 13:47

1 Answer 1


I've gone through Rosen's book cover to cover. If I remember correctly, he does not do the best job at setting the reader up for some exercises in the beginning.

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} $

you can also derive it via conditional proof

$ \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} $


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