# $(r \wedge \neg s) \rightarrow \neg q$

How can this English sentence be translated into a logical expression?

In Kenneth Rosan, the answer to this following sentence

“You cannot ride the roller coaster if you are under 4 feet tall unless you are older than 16 years old.”

is given as,

$$(r \wedge \neg s) \rightarrow \neg q$$

Where,

q: “You can ride the roller coaster.”

r: “You are under 4 feet tall.”

s: “ You are older than 16 years old.”

My solution:

So, I broke down this compound sentence as follows:

“[You cannot ride the roller coaster] if [you are under 4 feet tall] unless [you are older than 16 years old.]”

Now, substituting variables in given compound sentence.

($$\neg q$$) if (r unless s).

Applying equivalence formula for Q if P $$\Leftrightarrow$$ P $$\to$$ Q

(r unless s) $$\to$$ ($$\neg q$$)

Now, solving for unless. So, (r unless s) $$\Leftrightarrow$$ ($$\neg s \to r$$) ref.

($$\neg s \to r$$) $$\to$$ ($$\neg q$$)

Again solving for $$\to$$ (implication), we get:

(s $$\lor$$ r) $$\to$$ ($$\neg q$$)

So, my derivation is obviously wrong and does not match with Kennet Rosen.

My Question: What mistake I did? and How to derive the given answer systematically?

As noted by Jay, Kenneth Rosen interprets (r unless s) according to:

$$\begin{array}{cc|c} r & s & (r \text{ unless } s) \Leftrightarrow (\neg s \to r) \\\hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}$$

The issue turns out to be the order of operations for "unless".

You started by breaking it up like this

“[You cannot ride the roller coaster] if [you are under 4 feet tall] unless [you are older than 16 years old.]”

And substituted variables to get:

($$\neg q$$) if ($$r$$ unless $$s$$).

If instead we use a different order of operations to group these, it works out with your definition of unless. That is, we have:

$$((\neg q) \mathbf{\text{ if }} r) \mathbf{\text{ unless }} s$$

Now using $$(P \mathbf{\text{ if }} Q) \Leftrightarrow (Q \to P)$$

$$(r \to \neg q) \mathbf{\text{ unless }} s$$

Now using your $$(P \mathbf{\text{ unless }} Q) \Leftrightarrow (\neg P \to Q)$$

$$\neg s \to (r \to \neg q)$$

expanding

$$\neg s \to (\neg r \lor \neg q)$$

expanding

$$s \lor (\neg r \lor \neg q)$$

This is the same as the other result.
To see this, use associativity of logical or

$$(s \lor \neg r) \lor \neg q$$

Then turn it into an implication

$$\neg(s \lor \neg r) \to \neg q$$

Use demorgan's law

$$(\neg s \land r) \to \neg q$$

(EDIT: Previously I arrived at the answer with the same order of operations, but a different interpretation of unless: $$(r \text{ unless } s) = (r \text{ and } \neg s)$$. Because "(anything) unless True = True" seriously sounds wrong to me. My interpretation of unless worked in this case, but apparently is not the correct english interpretation. Apologies.)

• For this reason, we are going to translate ‘P unless Q’ with just ¬Q → P unless stated otherwise from rpi.edu – Ubi hatt Apr 19 '20 at 4:14
• Also, what formula you assumed for "r unless s"? – Ubi hatt Apr 19 '20 at 4:15
• @Ubihatt I used no formula. I filled it out how I interpret the phrase in english. So it appears this may just be a natural language ambiguity if you interpret it otherwise. – HBrown Apr 19 '20 at 4:17
• Kenneth Rosen also has formula for unless which exactly matches with formula embedded in my previous comment. Please check. Unless is a keyword. – Ubi hatt Apr 19 '20 at 4:19
• Interpreting "P unless Q" = "¬Q → P", would mean "You cannot ride the roller coaster if you are under 4 feet tall unless you are older than 16 years old." says you cannot ride the roller coaster if you are older than 16 years old, no matter what your height is. Is that really how you interpret that phrase? – HBrown Apr 19 '20 at 4:27

The mistake is to interpret the sentence as $$(\neg q)\ \textbf{if}\ (r\ \textbf{unless}\ s)$$ The correct interpretation is $$(\neg q\ \textbf{if}\ r)\ \textbf{unless}\ s$$

• You beat me to the punch =) Nicely done. – Jay Dunivin Apr 19 '20 at 5:01

I have a discrete math book by Kenneth Rosen, and here is an excerpt from the book listing the equivalent ways to express $$p \implies q$$ in English. The one you mentioned is boxed in blue.

But where did you go wrong? I believe it was at this step

Now, substituting variables in given compound sentence. ($$\neg q$$) if (r unless s).

The word unless does not attach $$s$$ to $$r$$; it attaches $$s$$ with the proposition $$\neg q$$ if $$r$$. The reason for this is that there was a proposition established prior to unless, which was the statement, "You cannot ride the roller coaster if you are under 4 feet tall." That sentence is established as a proposition; it is the $$q$$ in the table I provided.

So the correct statement using your scheme is

$$\big( \, \neg q \, \textbf{ if } \, r \, \big) \, \textbf{ unless } \, s,$$ which is equivalent to $$\neg s \implies (r \implies \neg q).$$ Can you take it from here?

• As you saw, I initially interpreted unless differently, and just happened to get the right answer erroneously. So I learned something here too. But I'm struggling how "True unless True" evaluates to True. "The dog is angry unless the dog is fed". If the dog is angry and the dog is fed, I would interpret that as false. Is my english screwed up, or what am I missing? – HBrown Apr 19 '20 at 5:43
• Maybe what you're missing is that $F \implies T$ is true (similarly, $F \implies F$ is true). "The dog is angry unless the dog is fed," which is the same as "If the dog is not fed, then the dog is angry." If the dog is indeed fed, then the statement "the dog is not fed" is false, of course. But then the statement "If the dog is not fed, then the dog is angry" is actually a true statement (a false hypothesis can be used to derive any conclusion). The only way the statement could be false is when the dog is not fed, but the dog is not angry. Does that help? – Jay Dunivin Apr 19 '20 at 5:49
• To add to what I said, "True unless True" is equivalent to "False implies True," which is a true statement, per the definition of "implies" that we have agreed to adopt. We have agreed that $p \implies q$ is false if and only if $p$ is true and $q$ is false, and $p \implies q$ is true otherwise. – Jay Dunivin Apr 19 '20 at 5:53
• I see where you're coming from. I don't think it's your English. I was looking at other sites and papers, and there isn't a consensus as to what "unless" means in general, for it can be construed in different ways in different contexts. This link is particularly helpful as to why in logic this particular convention was adopted: math.stackexchange.com/questions/1295106/… – Jay Dunivin Apr 19 '20 at 6:14
• You might also be interested in these other links regarding the ambiguity of "unless" outside of mathematics: Reddit: reddit.com/r/askphilosophy/comments/52585m/… and a linguistic paper: web.stanford.edu/~danlass/Nadathur-Lassiter-unless-SuB.pdf – Jay Dunivin Apr 19 '20 at 6:15