Elementary Logic
Build and check truth tables, translate compound propositions, and test logical equivalence.
Learning goals
- Recognize propositions and their primitive components.
- Construct truth tables using logical operators.
- Test equivalence and identify related conditionals.
Propositions and primitive components
Before deciding whether an argument works, we must identify the statements whose truth it uses. A proposition has a definite truth value, even when we do not know that value. We will separate these statements into primitive components so their combinations can be studied systematically.
A proposition is a declarative sentence with exactly one truth value: true or false.
Questions and commands are not propositions. An open sentence with an unassigned variable needs a value or a quantifier to determine a truth value.
Identify a proposition
Show worked solution
It is a proposition and is false: the digit sum is \(4\), not a multiple of three.
A compound proposition joins primitive propositions such as \(p,q,r,s\). The same letter must represent the same statement wherever it occurs.
A question, an open sentence, and a false statement
Classify “How are you doing?”, “x + 1 = 4”, and “Cebu is the capital of the Philippines.”
Show worked solution
The question has no truth value. The open sentence needs an assigned value or a quantifier. The last sentence is a proposition even though it is false: Manila is the capital. Being false does not stop a statement from being a proposition.
The five logical operators
We have identified the basic propositions. Now the connecting words determine how their truth values combine. Read each operator carefully instead of importing every everyday meaning of its word. In particular, a conditional is false when its premise is true and its conclusion is false; keep that specific case in mind while completing the tables.
| Name | Notation | Rule |
|---|---|---|
| Negation | \(\neg p\) | Reverse the truth value |
| Conjunction | \(p\land q\) | True only when both are true |
| Inclusive disjunction | \(p\lor q\) | False only when both are false |
| Conditional | \(p\to q\) | False only when the antecedent is true and consequent false |
| Biconditional | \(p\leftrightarrow q\) | True when the two values agree |
| \(p\) | \(q\) | \(p\land q\) | \(p\lor q\) | \(p\to q\) | \(p\leftrightarrow q\) |
|---|---|---|---|---|---|
| \(T\) | \(T\) | \(T\) | \(T\) | \(T\) | \(T\) |
| \(T\) | \(F\) | \(F\) | \(T\) | \(F\) | \(F\) |
| \(F\) | \(T\) | \(F\) | \(T\) | \(T\) | \(F\) |
| \(F\) | \(F\) | \(F\) | \(F\) | \(T\) | \(T\) |
Logical “or” is inclusive. A conditional does not assert that its antecedent is true.
Negation reverses a truth value
Negate “\(144\) is a perfect square.”
Show worked solution
“\(144\) is not a perfect square.” The original is true because \(12^2=144\); its negation is false.
| \(p\) | \(\neg p\) |
|---|---|
| \(T\) | \(F\) |
| \(F\) | \(T\) |
Conjunction and inclusive disjunction
Let \(p\) mean “A triangle has three sides” and \(q\) mean “\(2020\) is divisible by \(3\).” Evaluate their conjunction and disjunction.
Show worked solution
Here \(p=T\) and \(q=F\).
\[p\land q=F,\qquad p\lor q=T\]The conjunction needs both claims. Inclusive “or” needs at least one; it also allows both.
When does an implication fail?
Consider “If it is raining hard, then the plants outside are wet.” What would make this claim false?
Show worked solution
Rain together with dry plants is its counterexample: true antecedent, false consequent. Wet plants without rain do not refute it; another cause could have made them wet. When the antecedent is false, the material conditional is true.
A biconditional requires both directions
Explain \(p\leftrightarrow q\) using two conditionals.
Show worked solution
When both components are true or both false, both conditionals hold. If just one component is true, one direction fails, so the biconditional is false.
Translate with care
Before constructing a table, identify the main operation in the sentence. Words such as only if and unless deserve particular attention because a small change in wording can reverse a conditional or alter a negation. Check your translation against a situation in which the original statement would be false.
Let \(p\) mean “It is raining” and \(q\) mean “The streets are wet.” Then “rain is sufficient for wet streets” and “wet streets are necessary for rain” both translate as \(p\to q\).
Negate a conditional
Show worked solution
It is raining and the streets are not wet. This is the one situation that makes the original implication false.
Not both
Show worked solution
She did not visit France, or she did not visit Italy, possibly both.
Necessary and sufficient conditions
Translate “Raining is sufficient for the home team to win.” Let p mean it rains and q mean the home team wins.
Show worked solution
The sufficient condition goes before the arrow. Equivalently, winning is necessary for raining in this stated implication. This is a translation exercise, not a claim that weather actually guarantees a win.
Constructing a truth table
Why list every possible case?
A truth table lists every assignment of truth values to the distinct primitive propositions and the resulting truth values of the compound propositions. With a given number of independent primitive propositions, the row count is two raised to that number.
It is useful when the actual facts are unknown or when we want a claim valid regardless of those facts. Intermediate columns show how each operation contributes, making the final column’s reasoning checkable.
We have named the logical operations; now we need a method that checks every possible case. A truth table is a systematic record of those cases. Build the primitive columns first, then evaluate smaller parts before combining them into the full proposition.
With \(n\) distinct primitives, use \(2^n\) rows. Begin with the primitive columns, then add intermediate operations in dependency order. Use the standard row order starting with all true: the last primitive alternates every row; the preceding one every two rows, and so on.
Precedence is negation, conjunction, disjunction, conditional, then biconditional. Parentheses override this order. In our activity, the compound proposition is supplied; you enter T or F throughout the table.
A two-variable example
Construct a truth table for the following proposition, including the intermediate operations.
\((p\lor q)\land\neg p\)Show worked solution
| \(p\) | \(q\) | \(p\lor q\) | \(\neg p\) | \((p\lor q)\land\neg p\) |
|---|---|---|---|---|
| \(T\) | \(T\) | \(T\) | \(F\) | \(F\) |
| \(T\) | \(F\) | \(T\) | \(F\) | \(F\) |
| \(F\) | \(T\) | \(T\) | \(T\) | \(T\) |
| \(F\) | \(F\) | \(F\) | \(T\) | \(F\) |
Evaluate the disjunction and negation before their conjunction. Only the third row makes the final expression true.
Truth-table workshop
Let us put the method into practice on a compound proposition. First list every combination of the primitive truth values without repeating or skipping a row. Then work outward from the innermost operation. If a final entry is wrong, trace that row through the intermediate columns; this shows where the reasoning changed instead of forcing you to restart the whole table.
Tautologies, contradictions, and contingencies
The completed final column now lets us classify the whole proposition. Do not decide from one row: look at all possible assignments. Always true, always false, and true only in some cases are different kinds of logical behavior, even when two propositions agree in the particular situation you first imagined.
A tautology is true in every row; a contradiction is false in every row. A contingency has both true and false rows.
Exhaustive alternatives
Classify the proposition as a tautology, contradiction, or contingency. Justify every possible case.
\(p\lor\neg p\)Show worked solution
For either value of p, one disjunct is true. It is a tautology.
Incompatible claims
Classify the proposition and explain why its two parts cannot both hold.
\(p\land\neg p\)Show worked solution
A value and its negation cannot both be true. It is a contradiction.
Inspect every row. A single true evaluation does not prove a tautology.
A statement that depends on the row
Classify \((p\land q)\lor\neg p\).
Show worked solution
| \(p\) | \(q\) | \(p\land q\) | \(\neg p\) | \((p\land q)\lor\neg p\) |
|---|---|---|---|---|
| \(T\) | \(T\) | \(T\) | \(F\) | \(T\) |
| \(T\) | \(F\) | \(F\) | \(F\) | \(F\) |
| \(F\) | \(T\) | \(F\) | \(T\) | \(T\) |
| \(F\) | \(F\) | \(F\) | \(T\) | \(T\) |
The final column contains both truth values. It is neither a tautology nor a contradiction; we call it a contingency.
Logical equivalence
Different wording, the same truth conditions
Two propositional formulas are logically equivalent if and only if they have the same truth value under every assignment to all primitive propositions appearing in either formula.
This allows one formula to replace another in reasoning. A single row where their values differ disproves equivalence.
A completed table tells us more than the truth value in one situation. We can compare two entire final columns to see whether the propositions agree in every case. That is a stronger claim than noticing that both happened to be true in one example.
Two expressions are logically equivalent when their final columns match in every row. Their biconditional is then a tautology.
\[p\to q\equiv\neg p\lor q\]\[\neg(p\lor q)\equiv\neg p\land\neg q\]\[\neg(p\land q)\equiv\neg p\lor\neg q\]The switcheroo law
Verify this logical equivalence by identifying exactly when each side is false.
\(p\to q\equiv\neg p\lor q\)Show worked solution
Both columns are false exactly when p is true and q is false. In the other three assignments both are true.
Compare rain and wet streets
Are “If it rains, the streets are wet” and “If the streets are dry, it is not raining” equivalent? Compare with “It rains, or the streets are dry.”
Show worked solution
Let \(p\) mean it rains and \(q\) mean the streets are wet. Interpret dry as not wet.
\[p\to q\equiv\neg q\to\neg p\]The third statement is \(p\lor\neg q\). For \(p=T,q=F\), the original conditional is false but the third statement is true. Thus the first two are equivalent and the third is not.
A valid argument is not the same as true premises
An argument is valid if there is no assignment making all its premises true and its conclusion false. It is sound if it is valid and its premises are actually true. In a truth table, inspect rows where every premise is true; other rows cannot refute validity. A contradiction among the premises makes an argument valid in classical logic, but never sound.
An argument with inconsistent premises
Consider premises \(p\) and \(\neg p\), with conclusion q. Is the argument valid? Does it establish that q is actually true?
Show worked solution
No truth-table row makes both premises true, so there is no counterexample row: the argument is valid by the stated definition. It cannot be sound because the premises cannot both be true. Validity guarantees preservation of truth when all premises are true; it does not certify those premises.
State what the variables range over
The quantifier \(\forall x\in D\) means “for every x in D”; \(\exists x\in D\) means “for at least one x in D.” Negation exchanges these: \(\neg\forall x\,P(x)\equiv\exists x\,\neg P(x)\) and \(\neg\exists x\,P(x)\equiv\forall x\,\neg P(x)\). Specify the same domain throughout. One counterexample defeats a universal claim; one example establishes an existential claim.
Order changes what must be found
Compare \(\forall x\in\mathbb R\ \exists y\in\mathbb R:\ y>x\) with \(\exists y\in\mathbb R\ \forall x\in\mathbb R:\ y>x\).
Show worked solution
The first is true: after x is given, choose \(y=x+1\). The second is false: a proposed single y must also exceed \(x=y\), which would require \(y>y\). The first allows the witness to depend on x; the second demands one witness for every x.
Negating “every student answered every question” gives “some student did not answer some question.” It does not say that nobody answered anything.
Converse, inverse, and contrapositive
Changing the order or the signs in a conditional can produce a different claim. Read each version in ordinary language before comparing its table. The goal is to understand which changes preserve meaning, rather than assume that similar wording means logical equivalence.
| Form | Expression |
|---|---|
| Original | \(p\to q\) |
| Converse | \(q\to p\) |
| Inverse | \(\neg p\to\neg q\) |
| Contrapositive | \(\neg q\to\neg p\) |
The original and contrapositive are equivalent. The converse and inverse are equivalent to each other, but generally not to the original.
A familiar conditional
Show worked solution
Converse: If I stay home, it snows tonight. Inverse: If it does not snow tonight, I will not stay home. Contrapositive: If I do not stay home, it does not snow tonight.
Practice and reasoning
Count rows
Show worked solution
There are four distinct primitives; repeated occurrences do not add rows.
Find a counterexample
Show worked solution
No. Set p true and q false. The first is false and the second true. One mismatching row is enough.
A false disjunction
Show worked solution
Both disjuncts must be false. Thus p is true and q is false.
Truth teller and liar
Show worked solution
Ask either guard: “Which path would the other guard say is safe?” Both point toward the dangerous path. Choose the other path. This relies on both guards knowing the routes and each other’s behavior.
Additional practice: negations and conditionals
Negate each proposition
Treat each sentence as a proposition in its stated context. Negate the whole statement, rather than each part independently.
- If it is raining hard, then the plants outside are wet.
- She did not visit France but she visited Italy.
- The number \(x\) is prime and it is divisible by \(2\).
- If \(x\) is divisible by \(4\), then \(x\) is composite.
- \(3\ne7\) or \(9\) is a prime number.
Show worked solution
- It is raining hard and the plants outside are not wet. Use \(\neg(p\to q)\equiv p\land\neg q\).
- She visited France or she did not visit Italy. Use \(\neg(\neg p\land q)\equiv p\lor\neg q\).
- \(x\) is not prime or is not divisible by \(2\). Negating a conjunction gives a disjunction.
- \(x\) is divisible by \(4\) and is not composite. A conditional fails only in this case.
- \(3=7\) and \(9\) is not prime. Negating a disjunction gives a conjunction; a negation need not be true.
Related conditionals
For each statement write its converse, inverse, and contrapositive, in that order.
- If \(x=5\), then \(2x=10\).
- Raining is sufficient for the home team to win.
- If \(x\) and \(y\) are rational, then \(x+y\) is rational.
Show worked solution
- Converse: If \(2x=10\), then \(x=5\). Inverse: If \(x\ne5\), then \(2x\ne10\). Contrapositive: If \(2x\ne10\), then \(x\ne5\).
- Converse: If the home team wins, then it is raining. Inverse: If it is not raining, then the home team does not win. Contrapositive: If the home team does not win, then it is not raining.
- Converse: If \(x+y\) is rational, then both \(x\) and \(y\) are rational. Inverse: If at least one of \(x\) and \(y\) is irrational, then \(x+y\) is irrational. Contrapositive: If \(x+y\) is irrational, then at least one of \(x\) and \(y\) is irrational. Only the contrapositive is guaranteed equivalent to the original.
Additional practice: mixed exercises
Propositions and their components
Decide whether each sentence has a definite truth value. Distinguish open sentences from quantified claims.
- How are you doing?
- This sentence has either five or it has ten words.
- Cebu is the capital of the Philippines.
- \(2020\) is divisible by \(3\).
- \(a^2+b^2=c^2\).
- If \(x\) is divisible by \(4\), then \(x\) is composite.
- I am lying.
- Ten is not a prime number.
Show worked solution
- This is a question, not a proposition: it does not assert something true or false.
- This is a false proposition. Counting the words gives \(11\), so neither alternative holds. It is compound: “this sentence has five words” or “this sentence has ten words.”
- A false, simple proposition: the capital is Manila.
- A false, simple proposition: the digit sum is \(4\), and \(2020=3(673)+1\).
- With no values or quantifiers specified, this is an open sentence. It is true for \(a=3,b=4,c=5\), but false for \(a=b=c=1\).
- The conditional has primitive components “\(x\) is divisible by \(4\)” and “\(x\) is composite.” As written, \(x\) is unspecified. With the explicit interpretation “for every positive integer \(x\),” it is a true proposition: \(x=4k\) with \(k\ge1\) has divisor \(2\) strictly between \(1\) and \(x\).
- Interpreted as a claim that this very sentence is false, it produces the liar paradox and cannot consistently receive T or F in elementary two-valued logic.
- A true compound proposition: it negates “\(10\) is prime.” Since \(10=2\cdot5\), that primitive claim is false.
Negation and quantifiers
Negate the entire statement, keeping its scope clear.
- \(144\) is a perfect square.
- I drink more than \(8\) glasses of water every day.
- Every composite positive integer has at least \(3\) positive factors.
Show worked solution
- Negation: “\(144\) is not a perfect square.” This negation is false because \(144=12^2\); forming a negation does not require it to be true.
- Negation: “On at least one day, I drink at most \(8\) glasses of water.” Negating “every day” changes it to “at least one day”; it does not say “at most eight every day.”
- Negation: “Some composite positive integer has fewer than \(3\) positive factors.” A universal claim is negated by an existential claim. The original is true: \(1\), a nontrivial divisor, and the number itself are distinct factors.
Translate and construct truth tables
Let \(p\) mean “\(\pi\) is irrational” and \(q\) mean “\(22/7\) is rational.” Translate between words and symbols, then construct each truth table. The formal table lists all assignments, even though these particular mathematical statements are both true.
- \(p\land q\)
- \(\neg p\lor\neg q\)
- “\(\pi\) is not irrational but \(22/7\) is rational.”
- “Either \(\pi\) is irrational and \(22/7\) is rational, or \(\pi\) is not irrational or \(22/7\) is not rational.”
Show worked solution
- “\(\pi\) is irrational and \(22/7\) is rational.” Conjunction requires both components to be true.
Truth table, starting with all primitive propositions true \(p\) \(q\) \(p\land q\) T T T T F F F T F F F F - “\(\pi\) is not irrational or \(22/7\) is not rational.” Negate each component, then take their inclusive disjunction.
Truth table, starting with all primitive propositions true \(p\) \(q\) \(\neg p\) \(\neg q\) \(\neg p\lor\neg q\) T T F F F T F F T T F T T F T F F T T T - “But” is conjunction, giving \(\neg p\land q\).
Truth table, starting with all primitive propositions true \(p\) \(q\) \(\neg p\) \(\neg p\land q\) T T F F T F F F F T T T F F T F - The formula is \((p\land q)\lor\neg p\lor\neg q\). When the conjunction fails, at least one negated component is true, so the formula is a tautology.
Truth table, starting with all primitive propositions true \(p\) \(q\) \(p\land q\) \(\neg p\lor\neg q\) \((p\land q)\lor\neg p\lor\neg q\) T T T F T T F F T T F T F T T F F F T T
Conditionals and biconditionals in symbols
For a positive integer \(x\), let \(p\): “\(x\) is even”; \(q\): “\(x\) is composite”; \(s\): “\(x\) has a positive factor other than \(1\) and itself”; \(t\): “\(x=2k+1\) for some integer \(k\)”; \(u\): “\(x=2\).” Translate, then construct the formal truth table using only the primitive variables that occur.
- If \(x\) is odd, then \(x=2k+1\) for some integer \(k\).
- If \(x\) is even but not equal to \(2\), then \(x\) is composite.
- \(t\to\neg u\)
- \((\neg u\land\neg q)\to\neg p\)
- \(x\) is odd if and only if \(x=2k+1\) for some integer \(k\).
- \(\neg q\leftrightarrow\neg s\)
- \(u\leftrightarrow(p\land\neg q)\)
Show worked solution
- \(\neg p\to t\).
Truth table, starting with all primitive propositions true \(p\) \(t\) \(\neg p\) \(\neg p\to t\) T T F T T F F T F T T T F F T F - \((p\land\neg u)\to q\).
Truth table, starting with all primitive propositions true \(p\) \(u\) \(q\) \(\neg u\) \(p\land\neg u\) \((p\land\neg u)\to q\) T T T F F T T T F F F T T F T T T T T F F T T F F T T F F T F T F F F T F F T T F T F F F T F T - If \(x=2k+1\) for some integer \(k\), then \(x\ne2\).
Truth table, starting with all primitive propositions true \(t\) \(u\) \(\neg u\) \(t\to\neg u\) T T F F T F T T F T F T F F T T - If \(x\ne2\) and \(x\) is not composite, then \(x\) is not even.
Truth table, starting with all primitive propositions true \(u\) \(q\) \(p\) \(\neg u\land\neg q\) \(\neg p\) \((\neg u\land\neg q)\to\neg p\) T T T F F T T T F F T T T F T F F T T F F F T T F T T F F T F T F F T T F F T T F F F F F T T T - \(\neg p\leftrightarrow t\).
Truth table, starting with all primitive propositions true \(p\) \(t\) \(\neg p\) \(\neg p\leftrightarrow t\) T T F F T F F T F T T T F F T F - \(x\) is not composite if and only if it has no positive factor other than \(1\) and itself.
Truth table, starting with all primitive propositions true \(q\) \(s\) \(\neg q\) \(\neg s\) \(\neg q\leftrightarrow\neg s\) T T F F T T F F T F F T T F F F F T T T - \(x=2\) if and only if \(x\) is even and not composite.
Truth table, starting with all primitive propositions true \(u\) \(p\) \(q\) \(p\land\neg q\) \(u\leftrightarrow(p\land\neg q)\) T T T F F T T F T T T F T F F T F F F F F T T F T F T F T F F F T F T F F F F T
Classify formulas and verify equivalences
Construct a truth table. A tautology has only T in its final column; a contradiction has only F.
- \((p\land q)\lor\neg p\)
- \(\neg p\to\neg(q\lor r)\)
- \((p\lor q)\lor\neg(p\land q)\)
- For tautology \(\tau\) and contradiction \(\phi\), verify \(p\lor\tau\), \(p\land\phi\), \(p\lor\neg p\), and \(p\land\neg p\).
- Prove \(\neg(p\lor q)\equiv\neg p\land\neg q\).
Show worked solution
- Neither: the final column contains both truth values.
Truth table, starting with all primitive propositions true \(p\) \(q\) \(p\land q\) \(\neg p\) \((p\land q)\lor\neg p\) T T T F T T F F F F F T F T T F F F T T - Neither. The conditional fails exactly when \(p\) is false and at least one of \(q,r\) is true.
Truth table, starting with all primitive propositions true \(p\) \(q\) \(r\) \(\neg p\) \(q\lor r\) \(\neg(q\lor r)\) \(\neg p\to\neg(q\lor r)\) T T T F T F T T T F F T F T T F T F T F T T F F F F T T F T T T T F F F T F T T F F F F T T T F F F F F T F T T - Tautology. If the first disjunction is false, both components are false and the negated conjunction is true.
Truth table, starting with all primitive propositions true \(p\) \(q\) \(p\lor q\) \(\neg(p\land q)\) \((p\lor q)\lor\neg(p\land q)\) T T T F T T F T T T F T T T T F F F T T - The disjunctions are tautologies; the conjunctions are contradictions.
Truth table, starting with all primitive propositions true \(p\) \(p\lor\tau\) \(p\land\phi\) \(p\lor\neg p\) \(p\land\neg p\) T T F T F F T F T F - The last two columns match in every row, proving De Morgan’s law.
Truth table, starting with all primitive propositions true \(p\) \(q\) \(p\lor q\) \(\neg(p\lor q)\) \(\neg p\land\neg q\) T T T F F T F T F F F T T F F F F F T T
Compare everyday conditionals
Write symbols first, and keep the scope of each negation clear.
- Write the converse, inverse, and contrapositive of “If it snows tonight, then I will stay at home.”
- For real \(x,y\) with \(xy\ge0\), write the converse, inverse, and contrapositive of “If \(x\ge0\) and \(y\ge0\), then \(x+y\ge2\sqrt{xy}\).”
- Are these equivalent? (1) If it rains, the streets are wet. (2) If the streets are dry, it does not rain. (3) Either it rains or the streets are dry. Treat dry as not wet.
- Compare: (1) If a cell is an animal cell, it has neither a cell wall nor chloroplasts. (2) If it has a cell wall or chloroplasts, it is not an animal cell. (3) It has neither a cell wall nor chloroplasts, or it is not an animal cell.
Show worked solution
- Converse: If I stay at home, then it snows tonight. Inverse: If it does not snow tonight, then I will not stay at home. Contrapositive: If I do not stay at home, then it does not snow tonight.
- Converse: If \(x+y\ge2\sqrt{xy}\), then \(x\ge0\) and \(y\ge0\). Inverse: If \(x<0\) or \(y<0\), then \(x+y<2\sqrt{xy}\). Contrapositive: If \(x+y<2\sqrt{xy}\), then \(x<0\) or \(y<0\). The stated domain keeps the square root real in every version.
- Let \(p\) mean rain and \(q\) mean wet. The formulas are \(p\to q\), \(\neg q\to\neg p\), and \(p\lor\neg q\). The first two are contrapositives and equivalent. The third is not equivalent: at \(p=\mathrm F,q=\mathrm T\), the first two are true and the third is false.
- Let \(a,w,c\) denote animal cell, cell wall, and chloroplasts. The formulas are \(a\to(\neg w\land\neg c)\), \((w\lor c)\to\neg a\), and \((\neg w\land\neg c)\lor\neg a\). All simplify to \(\neg a\lor(\neg w\land\neg c)\), so they are equivalent. “Neither” makes the scope of the negation explicit.
Concept checks: explain your choice
Check recognition, translation, related conditionals, and equivalence. A hint gives a next step; feedback explains your checked choice.
Takeaways and connections
- Work from primitive columns to the final operation.
- A conditional fails only for true followed by false.
- Use all rows for equivalence, or one mismatch to disprove it.
Reasoning, connections and deeper practice
Truth tables test every possible assignment, but a short explanation can also show why an argument works. Distinguish the truth of one conditional from the validity of using premises to reach a conclusion.
Reasoning · Diagnose a reversed implication
A classroom rule states: “If a submission is late, it receives a flag.” A submission has a flag. Must it be late? Translate the reasoning and supply a truth assignment showing whether the conclusion follows.
Hint
Can a submission receive a flag for a different reason while the stated rule stays true?
Show worked solution
Let \(p\) mean late and \(q\) mean flagged. The premises are \(p\to q\) and \(q\); the proposed conclusion is \(p\). Set \(p=F,q=T\). The conditional and the second premise are true while the conclusion is false. Thus the argument is invalid: affirming the consequent is not justified. The rule does not say that lateness is the only reason for a flag.
Advanced / Honors · Prove a chained argument
Prove that \([(p\to q)\land(q\to r)]\to(p\to r)\) is a tautology. Explain why the truth-table workshop would require eight rows and which assignments your reasoning covers.
Hint
An implication can fail only when its antecedent is true. Separate whether p is true or false.
Show worked solution
If either premise conditional is false, the outer antecedent is false and the outer implication is true. Otherwise both premise conditionals are true. When \(p=F\), \(p\to r\) is true. When \(p=T\), the first premise forces \(q=T\), and the second forces \(r=T\), so \(p\to r\) is again true. These cases exhaust all assignments. There are three distinct primitives and therefore \(2^3=8\) truth-table rows; repeated appearances do not create new primitive columns.
