Questions and Practice Problems on Propositions: Worked Truth Tables and Logic Traps

Proposition questions look short, but one reversed arrow, missing bracket or unchecked truth assignment can change the answer completely. Use the same solve order every time: decide whether the sentence is a proposition, define the atoms, translate the wording, evaluate or tabulate, and finish with a classification or counterexample. The same solve order handles a central eight-row formula and practice problems on equivalence, validity and satisfiability.

Related reading: proposition MCQs and logical operator MCQs.

Propositions and practice problems: identify the object before calculating

A proposition is a declarative sentence with one definite truth value under a fixed interpretation. 11 is prime is T, while 15 is even is F. Both are propositions because being false does not stop a sentence from having a truth value.

In contrast, x + 4 = 10 is an open sentence until x is fixed or quantified. Close the window is a command, and Is 11 prime? is a question. None is a proposition in its given form.

The p, q and r, then join them to make compound formulas. With n distinct atoms, a truth table needs 2^n assignments. Three atoms need 8 rows; four need 16. See the Propositional and Predicate Logic: Truth Tables and Quantifiers develops the connective and quantifier foundations; mixed proposition practice requires the solve order used below.

Translate if, only if and exactly when before touching a truth table

Use p for "the lab is open", q for "the tutor is present", and r for "the practice session runs".

• "The practice session runs only if the lab is open and the tutor is present" is r -> (p ∧ q).

• "If the lab is open and the tutor is present, the practice session runs" is (p ∧ q) -> r.

• "The practice session runs exactly when both prerequisites hold" is r <-> (p ∧ q).

• "At least one prerequisite is missing" is not(p ∧ q), equivalently not p ∨ not q.

At p=T, q=F, r=T: r -> (p ∧ q) = T -> F = F; (p ∧ q) -> r = F -> T = T; r <-> (p ∧ q) = T <-> F = F; and not(p ∧ q) = not F = T. The first two results differ, so reversing "only if" changes the formula. Use Implication and Biconditional Operators in Logic to isolate arrow direction before combining it with classification and validity.

Fully worked truth table: prove a case-based argument is a tautology

Classify

F = [(p ∨ q) ∧ (p -> r) ∧ (q -> r)] -> r.

Set A = (p ∨ q) ∧ (p -> r) ∧ (q -> r), then calculate:

At (p,q,r)=(T,F,T), all three parts of A are true, so A=T and T -> T=T. At (p,q,r)=(T,T,F), both promises to r fail, making A=F; the outer implication is F -> F=T because its antecedent is false.

All eight final entries are T, so F is a tautology. If at least one case occurs and each possible case implies r, then r follows.

Equivalence, validity and satisfiability need different kinds of evidence

The claim tells you how much evidence you must produce. For equivalence, compare every assignment. For rows (T,T), (T,F), (F,T), (F,F), both p -> q and not p ∨ q produce T,F,T,T. Therefore p -> q ≡ not p ∨ q. One matching row is not enough because equivalence requires matching outputs on every row.

Now test the argument p ∨ q, p, therefore not q. Set p=T, q=T. The first premise is T ∨ T=T, the second premise is p=T, and the conclusion is not q=F. The premises are true while the conclusion is false, so the argument is invalid. This error treats inclusive OR as if exactly one input could be true.

Finally, satisfy G=(p ∨ q) ∧ (not p ∨ r) ∧ not q. The last part fixes q=F. Then p ∨ q forces p=T, and not p ∨ r forces r=T. Thus (p,q,r)=(T,F,T) makes G=T. One true assignment proves satisfiability; one counterexample disproves validity or equivalence.

Four proposition practice problems to solve before reading the answer

Evaluate not(p ∧ q) ∨ (q -> r) at p=T,q=F,r=T.

Since p ∧ q=F, the working is not F ∨ (F -> T)=T ∨ T=T.

Classify H=(p ∧ q) ∧ not p.

Regroup it as (p ∧ not p) ∧ q=F ∧ q=F. It is a contradiction.

Find a satisfying assignment for (p ∨ q) ∧ not p.

Choose p=F,q=T. Then the formula is T ∧ T=T, so the assignment satisfies it.

How many rows are needed for (p ∧ q) -> (r ∨ s)?

There are four distinct atoms. Therefore the table needs 2^4=16 rows.

Match the method to the demand. Use direct substitution for one supplied valuation. Use a full table or valid algebra for classification and equivalence. For validity, hunt for premises T and conclusion F. For satisfiability, find one assignment that makes the complete formula T.

Proposition traps: repair the exact wrong move

• Converse trap: Over the domain {1,2,3}, let p(x) mean x is in {1,2} and q(x) mean x is in {1,2,3}. p -> q is true for every domain element, but the converse q -> p fails at x=3. The contrapositive not q -> not p remains equivalent to the original.

• De Morgan trap: not(p ∨ q) is not p ∧ not q, not not p ∨ not q. At p=T,q=F, the correct left side is not T=F; the wrong expression is F ∨ T=T.

• Classification trap: One true row does not prove a tautology, and one false row does not prove a contradiction. A mixed final column means contingency. Ordinary ∨ is inclusive, so T ∨ T=T.

• Bracket trap: Do not evaluate not p ∧ q as not(p ∧ q). At p=T,q=F, the first is F ∧ F=F, while the second is not F=T.

How proposition questions are tested and how to practise them

Stable question forms include identifying propositions, translating wording, filling truth-table columns, classifying or comparing formulas, and validating an argument with a proof or counterexample.

For a three-variable problem, try a 12-minute loop: spend 2 minutes defining atoms and brackets, 5 minutes building the 8-row table, 3 minutes checking decisive rows, and 2 minutes recording the exact error. Accuracy comes before compressing the method. Repeat the loop before trying to work faster.

Use GATE CS Exam Preparation for the wider route and Propositional and Predicate Logic MCQs: 12 Solved for a focused drill set. Our practice bank has over 25 questions on this subtopic, though that does not mean every one is an official PYQ.

Questions on propositions: the short version and next step

Use five steps: test whether the sentence is a proposition, define the atoms, translate faithfully, choose the right evidence, and state the final classification. Check all rows to prove a tautology; one false row refutes it. Premises T with conclusion F refute validity, while one true assignment proves satisfiability. Rebuild the eight-row table without looking, then use GATE Guidance by Sanchit Sir for the broader Discrete Mathematics sequence and guided problem practice.