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An argument is valid when no valuation makes every premise true and the conclusion false, and that is the same as its corresponding conditional being a tautology.
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 state what validity is, test an argument by crossing out the rows where a premise fails and checking the conclusion on what survives, give the column of an argument's corresponding conditional, distinguish validity from soundness and from the truth of the conclusion, and say why an argument with jointly unsatisfiable premises is valid.
You can build a truth table and decide whether a formula is a tautology. Validity is that machinery pointed at a group of formulas: some premises and a conclusion, tested together on every row at once.
An argument is a set of premises and a conclusion. It is valid when no valuation makes every premise true and the conclusion false, written $\Gamma \models \phi$ and read $\Gamma$ entails $\phi$. A countermodel is a valuation that does make them all true with the conclusion false. An argument is sound when it is valid and its premises are in fact true.
An argument is valid when there is no valuation on which every premise is true and the conclusion is false. That is the whole definition, and three things follow from reading it literally. First, validity is about the form: the subject matter of the atoms never enters, so an argument is valid or not whatever its letters stand for. Second, it is a conditional claim — it says what holds if the premises do — so a valid argument may have false premises, and adding the claim that they are true is soundness, a different and stronger thing. Third, only the rows on which every premise survives can matter, and the rest of the table can be ignored. The same fact can be written as a single formula: $\Gamma \models \phi$ exactly when the corresponding conditional, the premises conjoined arrowing to the conclusion, is a tautology. So validity, tautology and unsatisfiability are three views of one test.
Another way: steps
Another way: example
From $P \to Q$ and $\neg Q$ conclude $\neg P$. The second premise leaves the two rows with $Q$ false; the first then removes the one with $P$ true. On the single surviving row $\neg P$ holds, so the argument is valid.
The commonest error is judging an argument by its conclusion: an argument with a true conclusion can be invalid, and an argument with a false one can be valid. The second is treating validity as a claim that the premises hold; it never is, which is why soundness is a separate word. The third is stopping the hunt after a few rows — a countermodel found settles invalidity at once, but validity is a claim about every row and needs all of them checked or an argument that covers them all.
From $P \to Q$ and $Q$, conclude $P$. Take the row with $P$ false and $Q$ true.
Look for the mixed row.
Both premises are true there and the conclusion is false, so that row is a countermodel.
One row is the whole refutation.
On other rows the conclusion happens to be true. That is irrelevant: one bad row is enough.
Validity is about all rows.
From $P \vee Q$ and $\neg P$ conclude $Q$. The second premise deletes the top half of the table.
Cross out rows premise by premise.
In the bottom half, the disjunction fails on the last row, so one row survives: $P$ false, $Q$ true.
Two premises, one survivor.
$Q$ is true there, so there is no countermodel and the argument is valid.
Check the conclusion last.
The second premise leaves the two rows with $P$ false, and the first premise is satisfied on both of them.
On the row with $Q$ true the conclusion fails, so that row is a countermodel and the argument is invalid.
Premises: P -> Q and Q. Conclusion: P. Is the argument valid? If it is not, give a row that breaks it.
P -> Q
Q
∴ P
valid invalid — countermodel:
An argument's corresponding conditional has the premises conjoined as its antecedent and the conclusion as its consequent. Give the column of ((P -> Q) & ~Q) -> ~P over the four rows for $P$ and $Q$.
| P | Q | ((P -> Q) & ~Q) -> ~P |
|---|---|---|
An argument is valid. Which of these does that guarantee?
Mark every argument below that is valid.
This task has no paper form; do it on a device.
Match each term to what it means.
| One with no valuation making every premise true and the conclusion false | One for which at least one such valuation exists | A valuation making every premise true and the conclusion false | A valid argument whose premises are in fact true | |
|---|---|---|---|---|
| A valid argument | ||||
| An invalid argument | ||||
| A countermodel | ||||
| A sound argument |
Premises: P and ~P. Conclusion: Q. Is the argument valid? If it is not, give a row that breaks it.
P
~P
∴ Q
valid invalid — countermodel:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An argument has premises $P \to Q$ and $Q$, over the atoms $P$ and $Q$. On how many of the four rows are both premises true at once?
Answer:
You can test an argument for validity and say what the verdict does and does not claim. Say in your own words why an argument can be valid and yet establish nothing.
8. Your turn: from $P \to Q$ and $\neg P$, does $\neg Q$ follow?, step 2