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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.
| Term | What it means |
|---|---|
| Consistent set | A set whose members can all be true on one assignment. |
| Model of a set | An assignment satisfying every member of the set. |
| Inconsistent set | A set with no assignment making all its members true. |
| Jointly | On the same assignment, not on a different assignment for each claim. |
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 $\{P \to Q, \neg Q\} \models \neg P$.
P is satisfiable and ~P is satisfiable, but {P, ~P} is inconsistent. The first needs P true; the second needs P false. A consistency witness must work for every member at once. This also explains why checking premises in separate tables can mislead unless the rows are aligned.
Start with any atom forced by a premise. In {P, P -> Q, ~Q}, the first forces P true. The conditional then forces Q true, while the last premise forces Q false. No assignment can meet all three requirements. But every pair is consistent: dropping any one statement removes the conflict. Inconsistency may therefore depend on the whole set rather than on an obvious opposite pair.
Adding premises cannot restore consistency to an inconsistent set, because any model of the larger set would also have to satisfy the original members. Removing a premise can restore consistency, but it need not: a different conflict may remain. After proposing a repair, supply a full model of the repaired set. Consistency does not establish that the surviving claims are actually true.
Consistency asks whether the supplied statements can all hold together. It is not a vote about which statements seem reasonable, and it does not require the statements to express the same idea. P and Q can be jointly true while concerning unrelated matters. A model assigns truth values so that every member of the set is true on the same assignment. One such model proves consistency.
Begin with direct atomic assertions or denials because they impose immediate values. A premise P requires P true; a premise ~Q requires Q false. Record these requirements before evaluating the remaining compound premises. If a later condition demands the opposite value for an already fixed atom, you have found a conflict in that candidate assignment. If another branch remains available, investigate it before concluding the whole set is inconsistent.
For the set {P | Q, ~P}, the denial forces P false. The disjunction then requires Q true. Substitute P false and Q true back into both original formulas: P | Q is true, and ~P is true. This is a complete witness. The formulas need not each be tautologies; they only need to share one satisfying assignment.
Compare {P | Q, ~P, ~Q}. The denials force both atoms false, which falsifies the disjunction. There is no alternative assignment satisfying the denials, so the set is inconsistent. Explaining that exhaustion matters: the failure of an arbitrary chosen row would not have been enough. Here the premises force the very row that fails the remaining requirement.
An empty set of premises imposes no restrictions, so every assignment satisfies it in the usual classical semantics. Adding a tautology also imposes no new restriction because it is true on every assignment. Adding a contradiction has the opposite effect: no assignment can satisfy it, so any set containing it is inconsistent. These boundary cases follow from the same definition rather than requiring special practical interpretations.
Some inconsistencies appear only when several statements interact. Consider {P, P -> Q, ~Q}. The direct assertion fixes P true. The conditional then requires Q true, while the denial requires Q false. Each formula is individually satisfiable, and each pair is consistent, yet all three together are inconsistent. Pairwise compatibility is therefore not a sufficient test for joint consistency.
You can demonstrate the pairwise claim explicitly. For {P, P -> Q}, set both atoms true. For {P, ~Q}, set P true and Q false. For {P -> Q, ~Q}, set both false. Each pair has its own model. None of those assignments satisfies the full triple. This example shows why models of different subsets cannot simply be combined as though their values never conflict.
A truth table offers an exhaustive method. Create a column for every member of the set, using the same atomic rows. Mark a row as a common model only if every premise column is true there. If at least one row is marked, the set is consistent. If no row is marked in a complete table, it is inconsistent. The final conjunction of all members expresses exactly that common-model requirement.
When checking a proposed witness, do not stop after the easiest premise. A row may satisfy a long conditional but fail a short atomic assertion. Nor should you evaluate each premise using a different assignment that happens to help it. The word jointly is the essential restriction: one fixed set of atomic values must satisfy every statement at once.
Consistency does not establish that a story is plausible, likely, or actually true. It identifies logical compatibility within the declared representation. A collection of mutually compatible but false reports can be consistent. A real-world background fact could also make a formally available assignment impossible in the application; if that fact matters, add its representation to the premises and retest the enlarged set.
When a set is inconsistent, some claim must be rejected, revised, or reinterpreted if the account is to be accepted as jointly true. Logic can show which combinations conflict. It does not, from inconsistency alone, identify the particular report that is factually wrong. The three scheduling statements in this lesson allow different repairs, each with a different factual interpretation.
Removing statements can create new models because it removes restrictions. Adding statements cannot rescue an inconsistent set: any model of the larger collection would already have to satisfy the impossible original collection. This is a useful directional check when evaluating proposed repairs. An extra unrelated fact does not erase a contradiction among the existing requirements.
However, deleting one suspect sentence may leave another conflict. The set {P, ~P, Q, ~Q} remains inconsistent if you remove P, because Q and ~Q still conflict. A repair needs a fresh common-model check of all surviving statements. Finding one previously unnoticed conflict is a diagnosis, not yet a guarantee that the rest of the collection is coherent.
Replacing a statement is also different from merely deleting it. If you weaken P & Q to P, you permit assignments with Q false that the original excluded. If you replace it with P | Q, you permit more assignments still. These changes can restore consistency but also change the information conveyed. State the revision explicitly and justify it with the relevant evidence or policy decision instead of presenting it as a truth-preserving rewrite.
To document a repair, list the retained or revised statements, give a model, and calculate every statement under that model. Then distinguish the formal result from the factual recommendation. 'This revised set is consistent' means its members can all be true together. 'This is the correct account of what happened' requires further evidence. A clear audit preserves both the usefulness and the limits of the consistency check.
A club writes three statements about one evening: the room is booked; if the room is booked, a supervisor is present; no supervisor is present. Let B mean booked and S mean supervisor present. The set is {B, B -> S, ~S}. The first statement makes B true. With that value, the conditional can be true only if S is true. The third statement requires S false. There is no shared assignment, so the three statements are inconsistent.
The diagnosis does not identify which statement is wrong. Perhaps the booking was canceled, perhaps a supervisor is available after all, or perhaps the conditional policy was misstated. Logic shows that all three cannot be true together within the declared reading. Evidence is needed to choose the repair.
Consider removing the booking claim. B false and S false now satisfy B -> S and ~S together, so the reduced set is consistent. Removing the no-supervisor claim instead leaves B true and S true as a model. Removing the conditional permits B true and S false. Each repair changes a different part of the account.
It would be a mistake to call the original set consistent because each sentence sounds plausible on its own, or because different members of the club support different sentences. A model is a single possible arrangement of the facts, not a collection of votes. Write the proposed values beside every surviving statement and evaluate them all before announcing that a repair works.
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.
List every requirement.
{P | Q, ~P}
A witness must satisfy both statements together.
Apply the atomic denial.
P=F
The premise ~P must be true.
Satisfy the remaining disjunction.
Q=T
With P false, Q must make the disjunction true.
Check the original members.
P | Q=T; ~P=T
Both use the same assignment P=F,Q=T.
State the consistency result.
Consistent; model P=F,Q=T
One common model is sufficient.
State the full set.
{P, P -> Q, ~Q}
The test concerns all three members.
Apply the direct assertion.
P=T
The first premise must hold.
Apply the conditional requirement.
Q=T
A true antecedent forces the consequent for the conditional to hold.
Apply the stated denial.
Q=F
The final premise denies that same Q.
Compare the forced values.
Q=T and Q=F conflict
One assignment cannot satisfy both demands.
State the conclusion.
Inconsistent
Every candidate satisfying the first two violates the last.
List the original collection.
{P, ~P, Q, ~Q}
Both atom pairs impose opposing values.
Make the proposed deletion.
Remove P
This changes the set rather than proving P false.
List the surviving requirements.
{~P, Q, ~Q}
All remaining members must still be checked.
Apply the first surviving denial.
P=F
This satisfies ~P without addressing Q.
Inspect the other two demands.
Q=T and Q=F
They still conflict, so this repair is insufficient.
Consider an explicitly different repair.
Retain {~P, Q}
Removing ~Q as well eliminates the remaining conflict.
Verify a model of that revision.
P=F,Q=T gives ~P=T,Q=T
This proves consistency of the revision without deciding which original reports were false.
Take the candidate assignment.
P=T,Q=F
These values are proposed for the entire set {P,~Q}.
Evaluate each member.
P=T; ~Q=T
The denial of false Q is true.
Report the common model.
Test joint consistency of {P | Q, P -> Q}. In PQ row order TT,TF,FT,FF, write T for a common model and F otherwise as four comma-separated symbols. Then write the common-model count and result symbol: C for consistent (at least one model), I for inconsistent (no models).
Common-model flags: b0
Model count: b1
Result symbol: b2
Audit {P, P -> Q, ~Q}, then delete ~Q and recount common models over P,Q.
Apply all original requirements.
P=T forces Q=T, while ~Q forces Q=F; before models
The demands cannot coexist on a single assignment.
Check the revised set.
{P,P -> Q} has after model
P and Q must both be true.
Separate repair from factual evidence.
The revision is consistent
Logic has not established which original premise should actually be rejected.
Test joint consistency of {P <-> Q, P, ~Q}. In PQ row order TT,TF,FT,FF, write T for a common model and F otherwise as four comma-separated symbols. Then write the common-model count and result symbol: C for consistent (at least one model), I for inconsistent (no models).
Common-model flags: b0
Model count: b1
Result symbol: b2
Test joint consistency of {~P, ~Q}. In PQ row order TT,TF,FT,FF, write T for a common model and F otherwise as four comma-separated symbols. Then write the common-model count and result symbol: C for consistent (at least one model), I for inconsistent (no models).
Common-model flags: b0
Model count: b1
Result symbol: b2
B means room booked and S means supervisor present. Write the combined claim: the room is booked and no supervisor is present.
Answer:
Test joint consistency of {P -> Q, ~(~P | Q)}. In PQ row order TT,TF,FT,FF, write T for a common model and F otherwise as four comma-separated symbols. Then write the common-model count and result symbol: C for consistent (at least one model), I for inconsistent (no models).
Common-model flags: b0
Model count: b1
Result symbol: b2
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Test joint consistency of {P -> Q, Q -> ~P, P | Q}. In PQ row order TT,TF,FT,FF, write T for a common model and F otherwise as four comma-separated symbols. Then write the common-model count and result symbol: C for consistent (at least one model), I for inconsistent (no models).
Common-model flags: b0
Model count: b1
Result symbol: b2
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.
14. Check a complete witness, step 3
Consistent
Both members hold on the same row.