Back to the on-screen lesson ·
Working backwards from a false conclusion to the one row that refutes an argument, and why a failed search that closes every route is a proof of validity.
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 construct a countermodel by falsifying the conclusion and satisfying the premises, fill in the whole table of an argument and read the breaking row off it, count the countermodels an argument has, refute an argument over three atoms without writing eight rows, and assign a value to an atom that occurs only in the conclusion.
You know that an argument is invalid exactly when some valuation makes every premise true and the conclusion false. This lesson is about producing that valuation rather than waiting for a table to reveal it.
A countermodel is a valuation making every premise true and the conclusion false. To refute an argument is to give one. A valuation is a value for every atom of the argument, including any atom that occurs only in the conclusion.
A table finds a countermodel by exhaustion; with three atoms that is eight rows and with five it is thirty-two, so the practical method runs the other way. Start from what a countermodel has to do and work backwards through the connectives. Make the conclusion false: if it is a conditional, that forces the antecedent true and the consequent false; if it is a conjunction, one conjunct must fail and there is a choice; if it is a disjunction, both must fail and there is none. Then take the premises one at a time and see whether they can still be satisfied with the atoms that are already fixed. If every route closes with a contradiction, the argument is valid, and the closed routes are the proof. If a route stays open, fill in the atoms it left free, in any way at all, and you have a countermodel. One is enough; there is nothing to gain by finding a second.
Another way: steps
Another way: example
From $P \to Q$ and $\neg P$, conclude $\neg Q$. The conclusion is false only with $Q$ true; the second premise needs $P$ false; the first is then satisfied. Countermodel: $P$ false, $Q$ true.
The first error is offering a row that makes the conclusion false without checking the premises on it; such a row refutes nothing, because the argument never promised anything there. The second is leaving an atom unassigned — most often one that appears only in the conclusion — which leaves a half-built valuation rather than a countermodel. The third is concluding validity from a failed search that was never exhaustive: the search must close every route, not merely the first two tried.
From $P \to Q$ and $Q \to R$, conclude $R \to P$. Make the conclusion false: $R$ true, $P$ false.
The conclusion first.
$P$ false satisfies the first premise outright. The second needs $Q \to R$, and $R$ is already true, so any $Q$ will do.
See what is forced and what is free.
Take $Q$ true. Countermodel: $P$ false, $Q$ true, $R$ true — one row instead of eight.
Fill the free atoms in.
From $P \to Q$ and $Q \to R$, conclude $P \to R$. Make the conclusion false: $P$ true, $R$ false.
Same opening move.
The first premise with $P$ true forces $Q$ true; the second with $Q$ true forces $R$ true.
Follow what is forced.
But $R$ was already false. The route closes, and it was the only route, so the argument is valid.
A closed search is a proof.
Make the conclusion false, so $P$ is false. The premise then needs $Q$ true.
Nothing contradicts that, so $P$ false with $Q$ true is a countermodel and the argument is invalid.
Premises: Q -> P and ~Q. Conclusion: ~P. Is the argument valid? If it is not, give the row that breaks it.
Q -> P
~Q
∴ ~P
valid invalid — countermodel:
The atom columns are filled in for you. Complete the columns for the two premises $P \to Q$ and $Q$ and for the conclusion $P$.
| $P$ | $Q$ | $P \to Q$ | $Q$ | $P$ | |
|---|---|---|---|---|---|
| Row 1 | T | T | |||
| Row 2 | T | F | |||
| Row 3 | F | T | |||
| Row 4 | F | F |
The argument is: from $P \to Q$ and $\neg P$, conclude $\neg Q$. Which valuation is a countermodel?
The argument is: from $P \to Q$, conclude $P$. Mark every row that is a countermodel.
This task has no paper form; do it on a device.
From the premise $P \to Q$, conclude $P$. Over the four rows for $P$ and $Q$, how many countermodels does this argument have?
Answer:
Premises: P -> Q and P. Conclusion: Q & R. Is the argument valid? If it is not, give a row that breaks it.
P -> Q
P
∴ Q & R
valid invalid — countermodel:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Premises: P -> Q and Q -> R. Conclusion: R -> P. Is the argument valid? If it is not, give a row that breaks it.
P -> Q
Q -> R
∴ R -> P
valid invalid — countermodel:
You can produce a countermodel for an invalid argument and explain why a search that closes every route establishes validity. Say in your own words why a row that makes the conclusion false is not yet a countermodel.
8. Your turn: from $P \vee Q$, does $P$ follow?, step 2