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

logicpropositional-calculus

Answers provided at this link do not satify my question.

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?

Best Answer

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.)

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