Determining Argument Validity — Worked Examples
Worked examples applying the methods from valid-argument-forms — truth table, rules of inference, and the invalidity/counterexample search — across every argument from Lecture 4, on both valid and invalid arguments.
The argument
Consider the argument:
If today is Monday, then I am wearing a pink shirt. Today is Monday. Therefore, I am wearing a pink shirt.
Let \(p\) = “today is Monday” and \(q\) = “I am wearing a pink shirt”. In symbolic form:
\[p \to q, \qquad p \qquad \therefore q\]
This is the Modus Ponens form (see valid-argument-forms).
Method 1: checking rows of the truth table
Build a truth table for the premises (\(p \to q\) and \(p\)):
| \(p\) | \(q\) | \(p \to q\) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
An argument is valid iff every row where all the premises are true also has the conclusion true. The premises here are \(p \to q\) and \(p\) — both true only in row 1 (\(p\)=T, \(q\)=T, \(p \to q\)=T). In that row the conclusion \(q\) is also T, and there’s no other row where both premises hold. So the argument is valid.
(To show an argument is invalid by this method, you instead go looking for a row where every premise is T but the conclusion is F — that’s the method used later in this lecture’s Activity 2(b)/3.)
Method 2: checking the combined statement is a tautology
Equivalently, build a truth table for \(\big((p \to q) \land p\big) \to q\) — the premises conjoined, implying the conclusion:
| \(p\) | \(q\) | \(p \to q\) | \(\big((p \to q) \land p\big) \to q\) |
|---|---|---|---|
| T | T | T | T |
| T | F | F | T |
| F | T | T | T |
| F | F | T | T |
Every row is T, so the statement is a tautology — the argument is valid. This is the same conclusion as Method 1, just packaged as one compound formula instead of a row-by-row check across separate premise/conclusion columns.
A valid argument can still have a false conclusion
This argument is valid regardless of whether \(p\) and \(q\) are actually true today — validity only says “if the premises hold, the conclusion follows”, not that the premises actually do hold. The premise “if today is Monday, then I am wearing a pink shirt” might well be false in reality, and the argument built from it is still valid: the form guarantees the conclusion given the premises, not the truth of the premises themselves.
Worked example 2: a three-variable argument
Consider the argument (Lecture 4, Activity 1):
If wages are raised, buying increases. If there is a depression, buying does not increase. Therefore there is not a depression or wages are not raised.
Let \(w\) = “wages are raised”, \(b\) = “buying increases”, \(d\) = “there is a depression”.
Translating to symbolic form
\[w \to b, \qquad d \to \sim b \qquad \therefore \sim d \lor \sim w\]
Truth table method
| \(w\) | \(b\) | \(d\) | \(w \to b\) | \(d \to \sim b\) | \(\sim d \lor \sim w\) |
|---|---|---|---|---|---|
| T | T | T | T | F | F |
| T | T | F | T | T | T |
| T | F | T | F | T | F |
| T | F | F | F | T | T |
| F | T | T | T | F | T |
| F | T | F | T | T | T |
| F | F | T | T | T | T |
| F | F | F | T | T | T |
Both premises (\(w \to b\) and \(d \to \sim b\)) are true together in rows 2, 6, 7, 8 — and the conclusion \(\sim d \lor \sim w\) is also T in every one of those rows. There’s no row where both premises hold and the conclusion fails, so the argument is valid.
Rules of inference method
- \(w \to b\) (premise 1)
- \(d \to \sim b\) (premise 2)
- \(\sim(\sim b) \to \sim d\) (contrapositive of 2)
- \(b \to \sim d\) (double negative law, from 3)
- \(w \to \sim d\) (Transitivity, from 1 and 4 — see valid-argument-forms)
- \(\sim w \lor \sim d\) (definition of \(\to\), from 5)
- \(\sim d \lor \sim w\) (commutativity, from 6) \(\blacksquare\)
Writing out the contrapositive step explicitly (line 3) before simplifying it (line 4) makes the justification for line 5’s Transitivity step airtight — Transitivity needs the forms \(w \to b\) and \(b \to \sim d\) to line up exactly.
Invalidity method, as a third check
Suppose, for contradiction, that all the premises are true but the conclusion is false.
- Conclusion false \(\Rightarrow\) \(\sim d \lor \sim w\) is false \(\Rightarrow\) \(\sim d\) is false and \(\sim w\) is false \(\Rightarrow\) \(d\) is true and \(w\) is true.
- Premise 1 (\(w \to b\)) is true and \(w\) is true \(\Rightarrow\) \(b\) is true.
- But now \(d\) is true and \(b\) is true (so \(\sim b\) is false), which makes premise 2 (\(d \to \sim b\)) false — contradicting the assumption that all the premises are true.
No consistent assignment exists, so the argument is valid — agreeing with both methods above.
All three methods (truth table, rules of inference, invalidity search) agree: this argument is valid.
Worked example 3: an argument via Conjunction, distribution, and Modus Ponens
Consider the argument (Lecture 4, Activity 2 — Q1 from the Methods video pre-class questions):
\[p \to q, \qquad p \lor r, \qquad p \lor \sim r \qquad \therefore q\]
Rules of inference method
- \(p \to q\) (given premise)
- \(p \lor r\) (given premise)
- \(p \lor \sim r\) (given premise)
- \((p \lor r) \land (p \lor \sim r)\) (from 2 and 3, Conjunction — see valid-argument-forms)
- \(p \lor (r \land \sim r)\) (from 4, distributive law — see logical-equivalence-laws)
- \(p \lor \mathbf{c}\) (from 5, negation law: \(r \land \sim r \equiv \mathbf{c}\))
- \(p\) (from 6, identity law)
- \(q\) (from 1 and 7, Modus Ponens) \(\blacksquare\)
The trick here is spotting that the two premises \(p \lor r\) and \(p \lor \sim r\) combine (via Conjunction, then distributing \(p\) out) into “\(p\), or a contradiction” — which forces \(p\) to be true regardless of \(r\).
Invalidity method, as a check
Suppose all the premises are true but the conclusion is false.
- Conclusion false \(\Rightarrow\) \(q\) is false.
- Premise 1 (\(p \to q\)) is true and \(q\) is false \(\Rightarrow\) \(p\) is false.
- Premise 2 (\(p \lor r\)) is true and \(p\) is false \(\Rightarrow\) \(r\) is true.
- Premise 3 (\(p \lor \sim r\)) is true and \(p\) is false \(\Rightarrow\) \(\sim r\) is true, i.e. \(r\) is false.
\(r\) can’t be both true and false — impossible. No such truth values exist, so the argument is valid, agreeing with the rules-of-inference proof above.
Worked example 4: an invalid argument
Not every argument is valid. Consider (Lecture 4, Activity 3 — Q2 from the Methods video pre-class questions):
\[p \to q, \qquad q \to r, \qquad \sim p \lor \sim q \qquad \therefore r\]
Invalidity method
Suppose all the premises are true but the conclusion is false.
- Conclusion false \(\Rightarrow\) \(r\) is false.
- Premise 2 (\(q \to r\)) is true and \(r\) is false \(\Rightarrow\) \(q\) is false.
- Premise 1 (\(p \to q\)) is true and \(q\) is false \(\Rightarrow\) \(p\) is false.
- Premise 3 (\(\sim p \lor \sim q\)), checked against \(p\) false and \(q\) false: it’s true. No contradiction so far.
This gives a consistent assignment — \(p\), \(q\), \(r\) all false — that makes every premise true and the conclusion false:
| \(p\) | \(q\) | \(r\) | \(p \to q\) | \(q \to r\) | \(\sim p \lor \sim q\) | \(r\) |
|---|---|---|---|---|---|---|
| F | F | F | T | T | T | F |
All three premises hold, the conclusion doesn’t — that’s a counterexample, so the argument is invalid.
Unlike the other arguments above, the search for a contradiction here succeeds in finding one instead — that’s what distinguishes an invalid argument from a valid one under this method.
See also
- valid-argument-forms
- logical-equivalence-laws
- practice-problems — §5 Methods for Determining Validity, with full worked solutions