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When a set of formulas has a model, what an inconsistent set entails, and how every entailment question becomes a consistency question.
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 count the models of a set of formulas, decide whether a set is consistent and produce a model when it is, test entailment from a set of premises, say what an inconsistent set entails and why, and convert an entailment question into a consistency question by adding the denial of the conclusion.
You can test one argument for validity and build a countermodel for it. This lesson looks at the premises on their own: what a set of formulas can be satisfied by, and what follows when nothing satisfies it.
A model of a set is a valuation making every member of it true. A set is consistent when it has at least one model and inconsistent when it has none. $\Gamma \models \phi$ says every model of $\Gamma$ is a model of $\phi$.
A set of formulas is consistent when some single valuation makes every member true at once. Satisfying each member separately is not enough; they must be satisfied together, and the work of testing is to try to build that one row. A set entails a formula when every model of the set is a model of the formula, which is the same validity test with the premises taken as a set. Two consequences are worth having in front of you. First, an inconsistent set entails everything: entailment asks that there be no valuation satisfying the set with the formula false, and a set with no models has no such valuation whatever the formula is. Second, and more useful, $\Gamma \models \phi$ exactly when $\Gamma \cup \{\neg \phi\}$ is inconsistent — so every entailment question can be turned into a consistency question and answered by trying to build a single row.
Another way: steps
Another way: example
$\{P \to Q, \neg Q, P\}$: the third member forces $P$ true, the first then forces $Q$ true, and the second denies it. No model, so the set is inconsistent — and therefore $\{P \to Q, \neg Q\} \models \neg P$.
The first error is checking the members one at a time and declaring the set consistent because each is satisfiable; they have to be satisfied on the same row. The second is reading inconsistency as saying that the members are false, when it says only that they cannot all be true together. The third is expecting an inconsistent set to entail nothing, when the definition makes it entail everything — which is exactly why inconsistency is worth detecting before anything is drawn from a set.
$\{P \vee Q, \neg P, Q \to R\}$: the second member fixes $P$ false.
Start where a value is forced.
The first then needs $Q$ true, and the third carries that to $R$ true.
Follow the forcing.
$P$ false, $Q$ true, $R$ true satisfies all three, so the set is consistent and that row proves it.
One row is the whole proof.
Does $\{P \vee Q, \neg P\}$ entail $Q$? Add $\neg Q$ to the set.
Deny the conclusion.
$\neg P$ and $\neg Q$ together deny both disjuncts of $P \vee Q$.
Look for a collision.
The enlarged set is inconsistent, so the entailment holds. One test does the work of two.
No model, so no countermodel.
The third member forces $P$ true, and the first then forces $Q$ true.
The fourth member denies $Q$, so the demands collide and the set has no model: it is inconsistent.
Over the four rows for $P$ and $Q$, how many satisfy all three of $P \to Q$, $Q \to P$ and $P$ at once?
Answer:
Mark every set below that is consistent.
This task has no paper form; do it on a device.
Premises: P -> Q, Q -> R and R. Conclusion: P. Does the conclusion follow? If it does not, give a row that breaks it.
P -> Q
Q -> R
R
∴ P
valid invalid — countermodel:
The set $\neg P$, $\neg Q$, $R$ has exactly one model. Give it: write T or F for each of the three atoms.
| $P$ | $Q$ | $R$ | |
|---|---|---|---|
| The model |
Match each term to what it means.
| Some valuation makes every member true | No valuation makes every member true | A valuation that makes every member true | Every valuation making every member true makes the formula true | |
|---|---|---|---|---|
| A consistent set | ||||
| An inconsistent set | ||||
| A model of a set | ||||
| A set entails a formula |
Premises: P -> (Q | R), ~Q and ~R. Conclusion: ~P. Is the argument valid? If it is not, give a row that breaks it.
P -> (Q | R)
~Q
~R
∴ ~P
valid invalid — countermodel:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A set of formulas is inconsistent. Which of these follows?
You can decide whether a set of formulas is consistent, build a model, and test entailment either directly or by denying the conclusion. Say in your own words why an inconsistent set entails every formula.
8. Your turn: is $\{P \to Q, Q \to P, P, \neg Q\}$ consistent?, step 2