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Modus ponens, modus tollens, disjunctive and hypothetical syllogism, the three fallacies they are confused with, and why a form applies only to whole formulas.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to name the standard valid forms and their schemas, recognise affirming the consequent, denying the antecedent and affirming a disjunct, say which row breaks each of them, write the conclusion a named form produces, and decide whether an argument is an instance of a form at all.
You can test an argument for validity and build a countermodel when it fails. This lesson collects the handful of patterns that come up constantly, so that the test need not be rerun every time, and names the near-misses that are the reason it sometimes must be.
A form is a pattern of premises and conclusion written with schematic letters. An instance of a form replaces those letters by formulas, uniformly. A formal fallacy is an invalid form, and the ones below are invalid because of their pattern rather than their subject matter.
Four patterns cover most of what is used. Modus ponens: from $\phi \to \psi$ and $\phi$, conclude $\psi$. Modus tollens: from $\phi \to \psi$ and $\neg \psi$, conclude $\neg \phi$. Disjunctive syllogism: from $\phi \vee \psi$ and $\neg \phi$, conclude $\psi$. Hypothetical syllogism: from $\phi \to \psi$ and $\psi \to \chi$, conclude $\phi \to \chi$. Each has a near-neighbour that fails. Affirming the consequent — from $\phi \to \psi$ and $\psi$, conclude $\phi$ — and denying the antecedent — from $\phi \to \psi$ and $\neg \phi$, conclude $\neg \psi$ — are both broken by the single row where the antecedent is false and the consequent true, and both amount to reading $\to$ as $\leftrightarrow$. Affirming a disjunct fails because $\vee$ allows both. A form applies only to instances: the letters stand for whole formulas, so affirming one conjunct of a compound antecedent is not affirming the antecedent.
Another way: steps
Another way: example
From $(P \vee Q) \to R$ and $\neg R$, conclude $\neg (P \vee Q)$. The consequent is denied and the whole antecedent is what gets negated, so this is modus tollens, and it is valid.
The commonest error is reading a conditional as a biconditional, which is what turns modus ponens into affirming the consequent and modus tollens into denying the antecedent. The second is matching a form to part of a formula: affirming $P$ when the antecedent is $P \wedge Q$ matches nothing. The third is treating a name as evidence — an argument is valid because no row breaks it, and the name is a record of that, not a substitute for it.
Affirming the consequent: from $P \to Q$ and $Q$, conclude $P$. Take $P$ false, $Q$ true.
The mixed row.
Denying the antecedent: from $P \to Q$ and $\neg P$, conclude $\neg Q$. The same row again.
One countermodel, two fallacies.
Both would be valid if $\to$ meant $\leftrightarrow$, which is exactly the row where those two differ.
The same mistake twice.
From $\neg P \to (Q \vee R)$ and $\neg P$, conclude $Q \vee R$.
Name the parts first.
$\phi$ is $\neg P$ and $\psi$ is $Q \vee R$; the second premise is exactly $\phi$.
Whole formulas, not letters.
So it is modus ponens and it is valid, even though every part of it is compound.
Instances may be as complicated as you like.
This affirms a disjunct rather than denying one, so it is not disjunctive syllogism.
The row with both atoms true keeps both premises true and makes $\neg P$ false, so the argument is invalid.
Match each named form to its schema.
| $P \to Q$ and $P$; therefore $Q$ | $P \to Q$ and $\neg Q$; therefore $\neg P$ | $P \vee Q$ and $\neg P$; therefore $Q$ | $P \to Q$ and $Q \to R$; therefore $P \to R$ | |
|---|---|---|---|---|
| Modus ponens | ||||
| Modus tollens | ||||
| Disjunctive syllogism | ||||
| Hypothetical syllogism |
Premises: P | Q and P. Conclusion: ~Q. Is the argument valid? If it is not, give a row that breaks it.
P | Q
P
∴ ~Q
valid invalid — countermodel:
Premises: P -> Q and P. Conclusion: Q. Which form is this?
For each form below, write T if it is valid and F if it is not.
| Valid? | |
|---|---|
| $P \to Q$, $P$; therefore $Q$ | |
| $P \to Q$, $\neg Q$; therefore $\neg P$ | |
| $P \to Q$, $Q$; therefore $P$ | |
| $P \to Q$, $\neg P$; therefore $\neg Q$ | |
| $P \vee Q$, $\neg P$; therefore $Q$ | |
| $P \vee Q$, $P$; therefore $\neg Q$ |
Mark every pattern below that is a valid form.
This task has no paper form; do it on a device.
Premises: (P & Q) -> R and P. Conclusion: R. Is the argument valid? If it is not, give a row that breaks it.
(P & Q) -> R
P
∴ R
valid invalid — countermodel:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
P stands for “the seal is intact” and Q for “the pressure holds”. From the premises P | Q and ~Q, write the conclusion that follows by disjunctive syllogism.
Answer:
You can name the form of an argument, say whether it is valid, and produce the row that breaks it when it is not. Say in your own words why affirming the consequent and denying the antecedent are broken by the same valuation.
8. Your turn: from $P \vee Q$ and $Q$, does $\neg P$ follow?, step 2