Forming Logic Expressions — Study Notes Summary & Study Notes

🧭 Steps to form logic expressions

Identify the atomic propositions: Break the situation into basic true/false statements (for example, “is a student”, “is 65 or older”, “order ≥ $20”). Give each a short variable name.

Choose a clear variable convention: Decide whether a variable is 1 for true or 1 for false. The usual convention is 1 = true, but always follow the problem’s table if it gives a different mapping.

Combine with logical operators: Use OR ($\lor$), AND ($\land$), and NOT ($\lnot$) to combine atomic propositions. Use parentheses to fix order of evaluation.

Translate the English condition into logic: Convert phrases like “either ... or ...” to $\lor$, “and” to $\land$, and negations like “not a student” to $\lnot$.

Simplify and check: Use Boolean identities (e.g., De Morgan’s laws, distributive/associative laws) to simplify. Always test a few cases or build a small truth table to verify the expression.

Canonical forms & tools: If needed, convert to DNF or CNF, or use a Karnaugh map for minimization.

✳️ Quick examples

• If $S$ = "is a student", $A$ = "is 65 or older", $M$ = "order ≥ $20": the condition “(student OR 65+) AND order ≥$20$” is

  $X = (S \lor A) \land M$.

• If a problem’s table defines a variable opposite to your preferred meaning (e.g., variable $V$ = 1 means "not a student"), express the intended proposition with a NOT: "student" = $\lnot V$.

✅ Common identities to use

• De Morgan: $\lnot (P \land Q) = (\lnot P) \lor (\lnot Q)$ and $\lnot (P \lor Q) = (\lnot P) \land (\lnot Q)$.
• Distributive: $P \land (Q \lor R) = (P \land Q) \lor (P \land R)$.
• Double negation: $\lnot(\lnot P) = P$.

Use these steps and rules each time you form or simplify a logic expression.

🧾 Problem statement (from screenshot)

Situation: The restaurant gives a discount ($X = 1$) if the person ordering is either a student or is 65 years or older, and the order amount is $20.00 or over.

The screenshot gives three rules with 0/1 encodings:

• Rule A: a student = 0, not a student = 1
• Rule B: under 65 years old = 0, 65 years or older = 1
• Rule C: under $20.00 = 0,$20.00 or over = 1

Because the table uses 1 for “not a student” for rule A, be careful with the meaning of that variable.

🔧 Define variables (matching the screenshot labels)

Let the given variables be:

• $A$: 1 means not a student, 0 means a student
• $B$: 1 means 65 years or older, 0 means under 65
• $C$: 1 means order $20.00 or over$, 0 means under $20.00$

The discount variable is $X$ where $X=1$ means discount given.

🧠 Form the logic expression

The English rule: "discount if (student OR 65+) AND amount ≥ $20". Using the screenshot variable meanings,$"student" is the negation ofA$, i.e.$\lnot A$. The age condition is$B$, and the amount condition is$C$. So the expression is:

$X = (\lnot A \lor B) \land C$

This reads: give discount when (student OR 65+) is true and order ≥ $20 is true.

🔍 Alternative if you prefer 1 = true for "student"

If you redefine variables so $S$ = "student" (1 = student), $G$ = "65+" (1 = 65+), $M$ = "order ≥ $20" (1 = yes), the expression is the simpler form:

$X = (S \lor G) \land M$

Both forms are equivalent once you map the variables consistently.

✔️ Quick truth-check

• A student with $C=1$ (order ≥ $20$): $A=0$, $B=0$, $C=1$ → $X = (\lnot 0 \lor 0) \land 1 = (1 \lor 0) \land 1 = 1$ (discount given).
• Non-student, 65+, with $C=1$: $A=1$, $B=1$, $C=1$ → $X = (\lnot 1 \lor 1) \land 1 = (0 \lor 1) \land 1 = 1$.
• Student with order under $20$: $C=0$ → $X = ... \land 0 = 0$ (no discount).

These checks confirm $X = (\lnot A \lor B) \land C$ is the correct translation of the screenshot problem.