Discrete Math Quiz: Truth Tables
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Truth TablesQuestion 1 of 19

When evaluating (p¬q)(¬pq)(¬p¬q)(p \land \neg q) \lor (\neg p \land q) \lor (\neg p \land \neg q) using a systematic truth table approach, the result is equivalent to which single logical operator applied to pp and qq?

¬(pq)\neg(p \lor q), which is the NOR operation between pp and qq
¬(pq)\neg(p \land q), which is the NAND operation between pp and qq
pqp \oplus q, which is the exclusive OR operation between pp and qq
pqp \leftrightarrow q, which is the biconditional operation between pp and qq
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Discrete Math Quiz

Discrete Math Quiz: Truth Tables

Practice Truth Tables in Discrete Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Truth Tables, giving you a quick way to practice the rules, question types, and explanations that matter most for Discrete Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

When evaluating (p¬q)(¬pq)(¬p¬q)(p \land \neg q) \lor (\neg p \land q) \lor (\neg p \land \neg q) using a systematic truth table approach, the result is equivalent to which single logical operator applied to pp and qq?

  1. ¬(pq)\neg(p \lor q), which is the NOR operation between pp and qq
  2. ¬(pq)\neg(p \land q), which is the NAND operation between pp and qq (correct answer)
  3. pqp \oplus q, which is the exclusive OR operation between pp and qq
  4. pqp \leftrightarrow q, which is the biconditional operation between pp and qq
Explanation: When you encounter complex logical expressions with multiple operators, the key is to systematically evaluate all possible truth value combinations and look for patterns that match simpler, equivalent expressions. Let's build a truth table for (p¬q)(¬pq)(¬p¬q)(p \land \neg q) \lor (\neg p \land q) \lor (\neg p \land \neg q):
pq¬p\neg p¬q\neg qp¬qp \land \neg q¬pq\neg p \land q¬p¬q\neg p \land \neg qFinal Result
TTFFFFFF
TFFTTFFT
FTTFFTFT
FFTTFFTT
The expression is true in three cases: when p is true and q is false, when p is false and q is true, and when both are false. It's only false when both p and q are true. Now compare this to ¬(pq)\neg(p \land q): this is false only when both p and q are true, and true in all other cases. This matches our result exactly, confirming answer B is correct. Option A (¬(pq)\neg(p \lor q)) is true only when both variables are false. Option C (pqp \oplus q) is false when both variables have the same truth value. Option D (pqp \leftrightarrow q) is true when both variables have the same truth value. Study tip: When simplifying complex logical expressions, always construct the complete truth table first, then compare the final column against standard logical operators. This systematic approach prevents errors and reveals equivalent forms.

Question 2

Using truth table evaluation, determine when the expression (pq)(qr)(rp)(p \leftrightarrow q) \land (q \leftrightarrow r) \land (r \leftrightarrow p) is true. How many of the 8 possible truth value assignments satisfy this condition?

  1. Exactly 4 assignments: when any two variables have the same truth value
  2. Exactly 2 assignments: when all three variables are true or all are false (correct answer)
  3. Exactly 6 assignments: when at least two variables have the same truth value
  4. Exactly 3 assignments: when exactly two variables have the same truth value
Explanation: When you encounter compound logical expressions with biconditionals (\leftrightarrow), remember that a biconditional pqp \leftrightarrow q is true only when both variables have the same truth value—either both true or both false. To solve this systematically, construct a truth table with all 8 possible combinations of truth values for pp, qq, and rr. For each row, evaluate each biconditional separately, then determine when all three are simultaneously true. Let's examine the key cases:
  • When p=T,q=T,r=Tp = T, q = T, r = T: All biconditionals are true (TT=TT \leftrightarrow T = T), so the entire expression is true.
  • When p=F,q=F,r=Fp = F, q = F, r = F: All biconditionals are true (FF=TF \leftrightarrow F = T), so the entire expression is true.
  • When variables have mixed values (like p=T,q=T,r=Fp = T, q = T, r = F): At least one biconditional becomes false (TF=FT \leftrightarrow F = F), making the entire conjunction false.
The pattern emerges: the expression is true only when all three variables have identical truth values. Answer A is incorrect because having any two variables the same doesn't guarantee the third matches. Answer C is wrong because "at least two the same" includes cases where the third differs. Answer D fails because "exactly two the same" means one variable differs, breaking at least one biconditional. Study tip: With multiple biconditionals connected by AND, look for cases where all variables are uniform. Biconditionals demand consistency—mixed truth values will always break the chain.

Question 3

Consider the statement: "If pp or qq is true, then both rr and ss are false." Using truth table analysis, under which conditions is this statement false?

  1. When pp and qq are both false, and either rr or ss is true
  2. When at least one of pp or qq is true, and at least one of rr or ss is true (correct answer)
  3. When pp and qq are both true, and rr and ss are both false
  4. When exactly one of pp or qq is true, and exactly one of rr or ss is false
Explanation: The statement translates to (pq)(¬r¬s)(p \lor q) \rightarrow (\neg r \land \neg s). An implication is false only when the antecedent is true and the consequent is false. The antecedent (pq)(p \lor q) is true when at least one of pp or qq is true. The consequent (¬r¬s)(\neg r \land \neg s) is false when at least one of rr or ss is true (since then not both rr and ss are false). Choice A describes when the antecedent is false, making the implication vacuously true. Choice C describes when both antecedent and consequent are true. Choice D is too restrictive about the "exactly one" conditions.

Question 4

Using a truth table to analyze (PQ)(¬P¬Q)(P \land Q) \leftrightarrow (\neg P \lor \neg Q), what type of logical expression is this?

  1. This expression is a tautology (always true)
  2. This expression is logically equivalent to PQP \oplus Q
  3. This expression is satisfiable but not a tautology
  4. This expression is a contradiction (always false) (correct answer)
Explanation: When analyzing compound logical expressions, constructing a truth table helps you determine whether the expression is always true (tautology), always false (contradiction), or sometimes true and sometimes false (satisfiable but not a tautology). Let's build the truth table for (PQ)(¬P¬Q)(P \land Q) \leftrightarrow (\neg P \lor \neg Q):
PQP ∧ Q¬P¬Q¬P ∨ ¬Q(P ∧ Q) ↔ (¬P ∨ ¬Q)
TTTFFFF
TFFFTTF
FTFTFTF
FFFTTTF
The final column shows all false values, making this a contradiction. The biconditional \leftrightarrow is true only when both sides have the same truth value, but (PQ)(P \land Q) and (¬P¬Q)(\neg P \lor \neg Q) always have opposite truth values. Answer A is wrong because the expression is never true. Answer B is incorrect because PQP \oplus Q (exclusive or) has the pattern F, T, T, F, not all false values. Answer C is wrong because a satisfiable expression must be true in at least one case, but this expression is false in every row. Study tip: Remember that De Morgan's Law tells us ¬(PQ)(¬P¬Q)\neg(P \land Q) \equiv (\neg P \lor \neg Q). So this expression is really asking whether (PQ)(P \land Q) equals its own negation, which is impossible—hence always false.

Question 5

A complex logical circuit evaluates ((PQ)(QR))(PR)((P \rightarrow Q) \land (Q \rightarrow R)) \rightarrow (P \rightarrow R). After constructing the complete truth table with variables PP, QQ, and RR, how many rows show this expression as false?

  1. Exactly two rows show the expression as false
  2. Exactly one row shows the expression as false
  3. No rows show the expression as false (correct answer)
  4. Exactly three rows show the expression as false
Explanation: This question tests your understanding of logical equivalences and the concept of tautologies. When you encounter a complex logical expression like this, you're essentially looking at a potential logical law that might always be true regardless of the truth values of its variables. The expression ((PQ)(QR))(PR)((P \rightarrow Q) \land (Q \rightarrow R)) \rightarrow (P \rightarrow R) represents the transitive property of implication. To verify this, construct a truth table with all eight possible combinations of P, Q, and R values. For each row, evaluate the antecedent (PQ)(QR)(P \rightarrow Q) \land (Q \rightarrow R) and the consequent (PR)(P \rightarrow R), then determine the truth value of the entire implication. When you work through each row systematically, you'll discover that this expression evaluates to true in every single case. This makes it a tautology—a logical statement that's always true regardless of the truth values of its components. The transitive property holds because whenever both "if P then Q" and "if Q then R" are true, "if P then R" must also be true. Answer choice A suggests exactly two false rows, which would indicate a partially valid logical relationship. Answer choice B proposes exactly one false row, suggesting the expression fails in just one scenario. Answer choice D claims exactly three false rows, indicating multiple logical failures. All of these are incorrect because the expression never evaluates to false. When studying logical equivalences, focus on recognizing fundamental logical laws like transitivity, which appear frequently in discrete mathematics. Tautologies are particularly important—they represent universally valid logical principles.

Question 6

A truth table is constructed for the expression ((PQ)R)(¬P¬Q¬R)((P \oplus Q) \land R) \lor (\neg P \land \neg Q \land \neg R). In how many rows does this expression evaluate to true?

  1. Exactly five rows evaluate to true
  2. Exactly four rows evaluate to true
  3. Exactly three rows evaluate to true (correct answer)
  4. Exactly six rows evaluate to true
Explanation: When you encounter a complex logical expression, the systematic approach is to construct a complete truth table with all possible combinations of the variables and evaluate the expression step by step. For ((PQ)R)(¬P¬Q¬R)((P \oplus Q) \land R) \lor (\neg P \land \neg Q \land \neg R), you need all eight possible combinations of P, Q, and R. Let's break this down systematically: The expression is true when either the first part ((PQ)R)((P \oplus Q) \land R) is true OR the second part (¬P¬Q¬R)(\neg P \land \neg Q \land \neg R) is true. Remember that PQP \oplus Q (exclusive or) is true when exactly one of P or Q is true, but not both. Working through each row:
  • The first part ((PQ)R)((P \oplus Q) \land R) is true when P and Q have different truth values AND R is true. This occurs in exactly two cases: (T,F,T) and (F,T,T).
  • The second part (¬P¬Q¬R)(\neg P \land \neg Q \land \neg R) is true only when all three variables are false: (F,F,F).
Since these conditions don't overlap, the expression evaluates to true in exactly three rows total. Answer A (five rows) likely comes from incorrectly counting cases where PQP \oplus Q is true regardless of R's value. Answer B (four rows) might result from misunderstanding the exclusive or operation. Answer D (six rows) could arise from treating \oplus as inclusive or rather than exclusive or. Strategy tip: For complex logical expressions, always work methodically through the truth table and double-check your understanding of each logical operator, especially \oplus which students often confuse with inclusive or.

Question 7

The expression ¬(PQ)(PQ)\neg(P \rightarrow Q) \land (P \lor Q) is evaluated using a truth table. Which truth value assignment makes this expression true?

  1. P=True,Q=TrueP = \text{True}, Q = \text{True}
  2. P=False,Q=FalseP = \text{False}, Q = \text{False}
  3. P=False,Q=TrueP = \text{False}, Q = \text{True}
  4. P=True,Q=FalseP = \text{True}, Q = \text{False} (correct answer)
Explanation: When you encounter compound logical expressions like this one, you need to systematically evaluate each component using the given truth values. This expression combines negation, implication, and conjunction, so understanding operator precedence and truth conditions is crucial. Let's work through option D: P=True,Q=FalseP = \text{True}, Q = \text{False}. First, evaluate PQP \rightarrow Q. An implication is false only when the antecedent is true and the consequent is false, which is exactly our case: TrueFalse=False\text{True} \rightarrow \text{False} = \text{False}. Therefore, ¬(PQ)=¬False=True\neg(P \rightarrow Q) = \neg\text{False} = \text{True}. Next, evaluate PQP \lor Q: TrueFalse=True\text{True} \lor \text{False} = \text{True}. Finally, the conjunction: TrueTrue=True\text{True} \land \text{True} = \text{True}. The entire expression is true. Option A fails because when both PP and QQ are true, PQP \rightarrow Q becomes TrueTrue=True\text{True} \rightarrow \text{True} = \text{True}, making ¬(PQ)=False\neg(P \rightarrow Q) = \text{False}, and FalseTrue=False\text{False} \land \text{True} = \text{False}. Option B fails because PQP \rightarrow Q becomes FalseFalse=True\text{False} \rightarrow \text{False} = \text{True}, so ¬(PQ)=False\neg(P \rightarrow Q) = \text{False}, and FalseFalse=False\text{False} \land \text{False} = \text{False}. Option C fails because PQP \rightarrow Q becomes FalseTrue=True\text{False} \rightarrow \text{True} = \text{True}, making ¬(PQ)=False\neg(P \rightarrow Q) = \text{False}, and FalseTrue=False\text{False} \land \text{True} = \text{False}. Remember: implications are only false when you have "true implies false." This makes option D the only case where negating the implication yields true, allowing the entire conjunction to be true.

Question 8

The logical expression ¬((pq)(¬q¬p))\neg((p \rightarrow q) \rightarrow (\neg q \rightarrow \neg p)) represents the negation of which fundamental logical principle?

  1. De Morgan's laws applied to conditional statements (though these concern negating conjunctions/disjunctions)
  2. The law of excluded middle in classical logic (though this involves different operators)
  3. The law of contraposition (creating a contradiction since contraposition is always valid) (correct answer)
  4. The distributive property of conjunction over disjunction (though this involves different logical structures)
Explanation: When you encounter logical expressions involving implications and their negations, you're typically dealing with fundamental principles of logical equivalence. The key here is recognizing what happens when you negate a well-established logical law. The expression (pq)(¬q¬p)(p \rightarrow q) \rightarrow (\neg q \rightarrow \neg p) represents the law of contraposition, which states that any conditional statement is logically equivalent to its contrapositive. This is always true in classical logic—contraposition is a fundamental principle that never fails. When you negate this expression with ¬((pq)(¬q¬p))\neg((p \rightarrow q) \rightarrow (\neg q \rightarrow \neg p)), you're creating a logical contradiction because you're asserting that contraposition doesn't hold, which is impossible. Let's examine why the other options don't fit: Option A mentions De Morgan's laws, but those specifically deal with negating conjunctions and disjunctions (¬(pq)¬p¬q\neg(p \land q) \equiv \neg p \lor \neg q), not the relationship between conditionals and their contrapositives. Option B refers to the law of excluded middle (p¬pp \lor \neg p), which involves disjunction and has a completely different logical structure. Option D discusses distributive properties between conjunction and disjunction, which again involves different operators and relationships than what we see in this implication-based expression. The correct answer is C because negating the law of contraposition creates an inherent logical contradiction. Study tip: When you see nested implications in logic problems, first identify if the inner structure represents a known logical principle. Contraposition problems often appear as implications between a conditional and its contrapositive form.

Question 9

Consider the nested conditional p(q(rs))p \rightarrow (q \rightarrow (r \rightarrow s)). This expression is logically equivalent to which of the following when expanded using truth table analysis?

  1. ¬p¬q¬rs\neg p \lor \neg q \lor \neg r \lor s, representing a disjunction of literals
  2. (pqr)s(p \land q \land r) \rightarrow s, representing a single conditional with compound antecedent
  3. ¬(pqr¬s)\neg(p \land q \land r \land \neg s), representing the negation of a specific conjunction
  4. All of the above expressions are logically equivalent to the original nested conditional (correct answer)
Explanation: Let's verify each equivalence. The original p(q(rs))p \rightarrow (q \rightarrow (r \rightarrow s)) can be rewritten as: ¬p(¬q(¬rs))=¬p¬q¬rs\neg p \lor (\neg q \lor (\neg r \lor s)) = \neg p \lor \neg q \lor \neg r \lor s (Choice A). This is also equivalent to (pqr)s(p \land q \land r) \rightarrow s because ¬p¬q¬rs=¬(pqr)s\neg p \lor \neg q \lor \neg r \lor s = \neg(p \land q \land r) \lor s (Choice B). By De Morgan's law, this equals ¬(pqr¬s)\neg(p \land q \land r \land \neg s) (Choice C). All three forms represent the same logical relationship: the nested conditional is false only when pp, qq, and rr are all true but ss is false. Each choice represents a valid equivalent form.

Question 10

In constructing a truth table for (pq)(qp)(p \rightarrow q) \oplus (q \rightarrow p), which pattern emerges?

  1. The expression is always true, forming a tautology in all cases
  2. The expression is true exactly when pp and qq have the same truth values
  3. The expression is always false, forming a contradiction in all cases
  4. The expression is true exactly when pp and qq have different truth values (correct answer)
Explanation: When you encounter logical expressions with exclusive or (\oplus) combined with conditionals, construct a systematic truth table to identify the pattern. Let's build the truth table for (pq)(qp)(p \rightarrow q) \oplus (q \rightarrow p): For p=T,q=Tp = T, q = T: Both pqp \rightarrow q and qpq \rightarrow p are true, so TT=FT \oplus T = F For p=T,q=Fp = T, q = F: pqp \rightarrow q is false, qpq \rightarrow p is true, so FT=TF \oplus T = T For p=F,q=Tp = F, q = T: pqp \rightarrow q is true, qpq \rightarrow p is false, so TF=TT \oplus F = T For p=F,q=Fp = F, q = F: Both conditionals are true, so TT=FT \oplus T = F The expression evaluates to true only when pp and qq have different truth values, confirming answer D. Answer A is wrong because the expression is false in two cases (when both variables match), so it's not a tautology. Answer B incorrectly states the opposite pattern—the expression is actually false when pp and qq have the same truth values. Answer C is wrong because the expression evaluates to true in two cases, so it's not a contradiction. Remember that exclusive or (\oplus) is true when its operands differ. When you see \oplus in logical expressions, look for patterns where the components have opposite truth values. This often reveals whether variables need to match or differ for the overall expression to be true.

Question 11

Consider the compound statement: (pq)(¬rs)(p \land q) \rightarrow (\neg r \lor s). If this statement is false, which of the following must be true?

  1. pp is true, qq is true, rr is true, and ss is false (correct answer)
  2. pp is false, qq is false, rr is false, and ss is true
  3. pp is true, qq is false, rr is true, and ss is false
  4. pp is false, qq is true, rr is false, and ss is true
Explanation: For an implication ABA \rightarrow B to be false, we need AA true and BB false. So (pq)(p \land q) must be true and (¬rs)(\neg r \lor s) must be false. For (pq)(p \land q) to be true, both pp and qq must be true. For (¬rs)(\neg r \lor s) to be false, both ¬r\neg r and ss must be false, meaning rr is true and ss is false. Choice B makes the antecedent false (since pp and qq are both false), which would make the implication true. Choice C makes the antecedent false (since qq is false). Choice D also makes the antecedent false (since pp is false).

Question 12

Given that the biconditional p(qr)p \leftrightarrow (q \land r) is true, and qsq \rightarrow s is false, what can be concluded about the truth value of psp \lor s?

  1. psp \lor s must be true in all valid cases
  2. psp \lor s must be false in all valid cases
  3. psp \lor s could be either true or false depending on other values (correct answer)
  4. psp \lor s is true when rr is true and false when rr is false
Explanation: From qsq \rightarrow s being false, we know qq is true and ss is false. From p(qr)p \leftrightarrow (q \land r) being true, pp and (qr)(q \land r) have the same truth value. Since qq is true, (qr)(q \land r) depends on rr. Case 1: If rr is true, then pp is true, so ps=TF=Tp \lor s = T \lor F = T. Case 2: If rr is false, then pp is false, so ps=FF=Fp \lor s = F \lor F = F. Therefore, psp \lor s can be either true or false. Choice A incorrectly assumes psp \lor s is always true. Choice B incorrectly assumes it's always false. Choice D correctly identifies the dependency on rr but states it backwards.

Question 13

Given the compound statement ((pq)r)(r(st))((p \lor q) \rightarrow r) \land (r \rightarrow (s \land t)), if pp is true and the entire compound statement is true, what must be true about ss and tt?

  1. ss and tt can have any truth values, depending on rr
  2. At least one of ss or tt must be true, but not necessarily both
  3. Both ss and tt must be true, regardless of other variable values (correct answer)
  4. ss and tt must be true only if qq is also true
Explanation: When analyzing compound logical statements, you need to trace through the implications systematically, especially when given specific conditions that make the entire statement true. Given that pp is true and the entire compound statement ((pq)r)(r(st))((p \lor q) \rightarrow r) \land (r \rightarrow (s \land t)) is true, let's work through what must follow. Since pp is true, (pq)(p \lor q) is automatically true regardless of qq's value. For the first part ((pq)r)((p \lor q) \rightarrow r) to be true when (pq)(p \lor q) is true, rr must also be true (since a true antecedent in a true implication forces a true consequent). Now that we know rr is true, look at the second part: (r(st))(r \rightarrow (s \land t)). Since rr is true and this implication must be true for the entire compound statement to be true, (st)(s \land t) must be true. For a conjunction to be true, both components must be true, so both ss and tt must be true. Choice A is wrong because rr is forced to be true, eliminating variability. Choice B misunderstands conjunctions—(st)(s \land t) requires both variables to be true, not just one. Choice D incorrectly suggests qq's value matters, but since pp alone makes (pq)(p \lor q) true, qq is irrelevant to the chain of reasoning. The answer is C: both ss and tt must be true regardless of other variables. Study tip: In logical chains, trace through each implication step-by-step. When the entire compound statement must be true, work forwards from your known values to find what else must be true.

Question 14

A student claims that ¬(pq)(¬p¬q)\neg(p \land q) \rightarrow (\neg p \lor \neg q) is a tautology. Using truth table analysis, what can be concluded?

  1. The student is correct; this is a tautology by De Morgan's law (correct answer)
  2. The student is incorrect; this is only true when pp and qq are both false
  3. The student is incorrect; this is a contradiction that's always false
  4. The student is incorrect; this is contingent, being false only when pp and qq are both true
Explanation: By De Morgan's law, ¬(pq)(¬p¬q)\neg(p \land q) \equiv (\neg p \lor \neg q). So the expression becomes (¬p¬q)(¬p¬q)(\neg p \lor \neg q) \rightarrow (\neg p \lor \neg q), which is of the form AAA \rightarrow A. This is always true regardless of the truth value of AA, making it a tautology. We can verify: when p=T,q=Tp=T, q=T: FF=TF \rightarrow F = T. When p=T,q=Fp=T, q=F: TT=TT \rightarrow T = T. When p=F,q=Tp=F, q=T: TT=TT \rightarrow T = T. When p=F,q=Fp=F, q=F: TT=TT \rightarrow T = T. Choice B suggests limited truth conditions. Choice C claims it's always false. Choice D claims it's sometimes false.

Question 15

The expression (pq)(qr)(rp)(p \rightarrow q) \land (q \rightarrow r) \land (r \rightarrow p) forms a logical cycle. In how many rows of its truth table does this expression evaluate to true?

  1. 2 rows: when all variables have the same truth value (correct answer)
  2. 4 rows: when any two variables have the same truth value
  3. 6 rows: when at least two variables have the same truth value
  4. 3 rows: when exactly two variables have the same truth value
Explanation: This expression is true when all three implications are true simultaneously. Since we have pqrpp \rightarrow q \rightarrow r \rightarrow p, this creates a cycle where all variables must have the same truth value. When p=q=r=Tp=q=r=T: (TT)(TT)(TT)=TTT=T(T \rightarrow T) \land (T \rightarrow T) \land (T \rightarrow T) = T \land T \land T = T. When p=q=r=Fp=q=r=F: (FF)(FF)(FF)=TTT=T(F \rightarrow F) \land (F \rightarrow F) \land (F \rightarrow F) = T \land T \land T = T. Any other combination breaks at least one implication. For example, if p=T,q=F,r=Tp=T, q=F, r=T, then pq=Fp \rightarrow q = F. Choice B incorrectly suggests 4 solutions. Choice C overcounts. Choice D undercounts and has wrong reasoning.

Question 16

Consider the truth table for the expression (PQ)(PR)(P \land Q) \rightarrow (P \lor R). This expression is false in exactly how many rows of the complete truth table?

  1. The expression is false in exactly zero rows (correct answer)
  2. The expression is false in exactly one row
  3. The expression is false in exactly two rows
  4. The expression is false in exactly three rows
Explanation: An implication ABA \rightarrow B is false only when AA is true and BB is false. Here, A=(PQ)A = (P \land Q) and B=(PR)B = (P \lor R). For the expression to be false, we need (PQ)=True(P \land Q) = \text{True} and (PR)=False(P \lor R) = \text{False}. If (PQ)=True(P \land Q) = \text{True}, then both PP and QQ are true. But if P=TrueP = \text{True}, then (PR)=True(P \lor R) = \text{True} regardless of RR's value. Therefore, it's impossible to have (PQ)=True(P \land Q) = \text{True} and (PR)=False(P \lor R) = \text{False} simultaneously. The expression is a tautology. Choices B, C, and D incorrectly assume there exist rows where the implication fails.

Question 17

Consider the compound statement: ((PQ)R)(¬RP)((P \land Q) \rightarrow R) \land (\neg R \land P). For which truth value assignment is this statement true?

  1. P=True,Q=True,R=FalseP = \text{True}, Q = \text{True}, R = \text{False}
  2. P=True,Q=False,R=FalseP = \text{True}, Q = \text{False}, R = \text{False} (correct answer)
  3. P=False,Q=True,R=FalseP = \text{False}, Q = \text{True}, R = \text{False}
  4. P=False,Q=False,R=TrueP = \text{False}, Q = \text{False}, R = \text{True}
Explanation: For the compound statement to be true, both conjuncts must be true: ((PQ)R)((P \land Q) \rightarrow R) and (¬RP)(\neg R \land P). From the second conjunct, we need ¬R\neg R true and PP true, so R=FalseR = \text{False} and P=TrueP = \text{True}. For the first conjunct with R=FalseR = \text{False}, we need (PQ)False(P \land Q) \rightarrow \text{False} to be true, which requires (PQ)=False(P \land Q) = \text{False}. Since P=TrueP = \text{True}, we need Q=FalseQ = \text{False}. Choice A fails because PQ=TrueP \land Q = \text{True} makes the implication false. Choice C fails because P=FalseP = \text{False} makes ¬RP\neg R \land P false. Choice D fails because R=TrueR = \text{True} makes ¬RP\neg R \land P false.

Question 18

A logic circuit implements the expression ¬((PQ)(¬P¬Q))\neg((P \land Q) \lor (\neg P \land \neg Q)). For which input combinations does this circuit output true?

  1. When PP and QQ have the same truth values
  2. When PP and QQ have different truth values (correct answer)
  3. When at least one of PP or QQ is false
  4. When both PP and QQ are true
Explanation: The expression (PQ)(¬P¬Q)(P \land Q) \lor (\neg P \land \neg Q) is true when either both P,QP,Q are true OR both are false (i.e., when they have the same truth value). The negation of this expression is therefore true when PP and QQ have different truth values. For P=T,Q=FP=T, Q=F: ¬((TF)(FT))=¬(FF)=¬F=T\neg((T \land F) \lor (F \land T)) = \neg(F \lor F) = \neg F = T. For P=F,Q=TP=F, Q=T: ¬((FT)(TF))=¬(FF)=T\neg((F \land T) \lor (T \land F)) = \neg(F \lor F) = T. Choice A describes when the non-negated expression is true. Choice C is too broad and includes cases where the output is false. Choice D is too specific and misses one of the true cases.

Question 19

Using the truth table for the expression (PQ)(QR)(¬R)(P \rightarrow Q) \land (Q \rightarrow R) \land (\neg R), determine which of the following must be true when the entire expression evaluates to true.

  1. Both PP and QQ must be false, and RR must be false (correct answer)
  2. PP must be false, QQ can be either true or false, and RR must be false
  3. PP can be either true or false, QQ must be false, and RR must be false
  4. PP must be true, QQ must be false, and RR must be false
Explanation: For the conjunction to be true, all three parts must be true: (PQ)(P \rightarrow Q), (QR)(Q \rightarrow R), and ¬R\neg R. From ¬R\neg R, we know R=FalseR = \text{False}. From (QR)(Q \rightarrow R) with R=FalseR = \text{False}, for this to be true we need Q=FalseQ = \text{False} (since TrueFalse=False\text{True} \rightarrow \text{False} = \text{False}). From (PQ)(P \rightarrow Q) with Q=FalseQ = \text{False}, for this to be true we need P=FalseP = \text{False}. Choice B incorrectly suggests QQ can be true. Choice C incorrectly suggests PP can be true. Choice D incorrectly states PP must be true, which would make PQP \rightarrow Q false when QQ is false.