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supply a finite countermodel to a stated quantified inference
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.
You will supply a finite countermodel to a stated quantified inference, recording the intermediate model values and the precise reason each conclusion follows.
Recall the truth conditions for not, and, or and if-then. Those rules still govern compound formulas here, but atomic truth now comes from objects and predicate extensions or from a world's valuation. Identify which new structure this lesson introduces before using a familiar propositional rule.
| Term | What it means |
|---|---|
| Interpretation | A declared domain and meanings for the nonlogical symbols; a modal interpretation also specifies worlds, accessibility and valuations. |
| Assignment | A choice of domain object for a free variable during an evaluation; it is not itself another domain object. |
| Witness | An eligible object or accessible world satisfying the property required by an existential or possibility claim. |
| Counterexample | An eligible case where the required condition fails; a countermodel to an inference additionally makes every premise true. |
| Validity | Truth in every interpretation of the specified kind, a stronger claim than truth in one supplied model. |
A first-order countermodel to an inference makes every premise true and the conclusion false in one interpretation. The same domain, denotations and predicate extensions must be used throughout. Showing a true premise in one model and a false conclusion in another is not a countermodel. A false conclusion alone is insufficient if a premise also fails.
For the inference from exists x P(x) and exists x Q(x) to exists x (P(x) and Q(x)), choose two distinct objects. Let only the first have P and only the second have Q. Both premises have witnesses, but no single object has both properties. The conjunction inside the conclusion requires the same witness for its two conjuncts. The premises never imposed that requirement.
This model establishes that the inference is invalid. It does not show that the conclusion is false in every model of the premises. Overlapping predicate extensions make the conclusion true. Validity asks for truth preservation in all models, so one countermodel defeats it even when many other interpretations happen to satisfy the conclusion.
A finite countermodel is decisive when found. Failure to find one in a few small domains is not a general proof of validity. First-order logic admits infinite interpretations, and some satisfiable theories have no finite models. The course's bounded searches therefore assess constructing and checking particular countermodels rather than pretending that a short finite search decides arbitrary first-order validity.
Another way: An explicit audit sheet
Keep four parts on the page: the declared objects or worlds, the meaning of each symbol, the intermediate values, and the conclusion. A changed interpretation belongs on a new sheet so the premises and conclusion are never checked in different models.
Domain: a, b, all distinct. P holds exactly of a; Q holds exactly of b. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
Object a: P is T and Q is F, so the conjunction is F. A conjunction needs both properties of this same object. A successful P-witness at one row cannot be combined with a Q-witness at a different row as though the two names denoted one object.
Object b: P is F and Q is T, so the conjunction is F. A conjunction needs both properties of this same object. A successful P-witness at one row cannot be combined with a Q-witness at a different row as though the two names denoted one object.
exists x P(x): T. The first extension contains its declared object, so the first premise has a witness. This is one of the requirements for a countermodel: every premise must be true in the same interpretation. Choosing an interpretation in which this premise is false would not refute the inference, regardless of what happened to the conclusion. Keep the domain fixed throughout the audit.
∃x Q(x): T. The second extension is also nonempty, so the second premise is true. Nothing in the two premises requires the witnesses to be identical. Distinct existential sentences can be witnessed by distinct objects. The case deliberately exploits that freedom without changing the meanings of P or Q between sentences. Both premise checks therefore belong to one model, not two separate examples.
exists x (P(x) and Q(x)): F. The two extensions are disjoint. Their intersection contains no object, so nobody witnesses the conjunction. With true premises and a false conclusion, this interpretation is a countermodel to the inference. It establishes invalidity, not that the conclusion is always false. In another model where the extensions overlap, the same conclusion could be true while the inference remains invalid in general.
Merge the two witness roles by putting the P-witness into Q as well as P. The conclusion becomes true, so that modified interpretation is no longer a countermodel. The original countermodel still establishes invalidity: one failed truth-preservation case suffices. This comparison distinguishes a countermodel from a mere model of the premises and prevents a common mistake of assuming that every interpretation must refute an invalid argument.
Domain: a, b, c, all distinct. P holds exactly of b; Q holds exactly of c. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
Object a: P is F and Q is F, so the conjunction is F. A conjunction needs both properties of this same object. A successful P-witness at one row cannot be combined with a Q-witness at a different row as though the two names denoted one object.
Object b: P is T and Q is F, so the conjunction is F. A conjunction needs both properties of this same object. A successful P-witness at one row cannot be combined with a Q-witness at a different row as though the two names denoted one object.
Object c: P is F and Q is T, so the conjunction is F. A conjunction needs both properties of this same object. A successful P-witness at one row cannot be combined with a Q-witness at a different row as though the two names denoted one object.
exists x P(x): T. The first extension contains its declared object, so the first premise has a witness.
∃x Q(x): T. The second extension is also nonempty, so the second premise is true.
exists x (P(x) and Q(x)): F. The two extensions are disjoint.
Merge the two witness roles by putting the P-witness into Q as well as P. The conclusion becomes true, so that modified interpretation is no longer a countermodel. The original countermodel still establishes invalidity: one failed truth-preservation case suffices. This comparison distinguishes a countermodel from a mere model of the premises and prevents a common mistake of assuming that every interpretation must refute an invalid argument.
Domain: a, b, c, d, all distinct. P holds exactly of c; Q holds exactly of d. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
Object a: P is F and Q is F, so the conjunction is F. A conjunction needs both properties of this same object. A successful P-witness at one row cannot be combined with a Q-witness at a different row as though the two names denoted one object.
Object b: P is F and Q is F, so the conjunction is F. A conjunction needs both properties of this same object. A successful P-witness at one row cannot be combined with a Q-witness at a different row as though the two names denoted one object.
Object c: P is T and Q is F, so the conjunction is F. A conjunction needs both properties of this same object. A successful P-witness at one row cannot be combined with a Q-witness at a different row as though the two names denoted one object.
Object d: P is F and Q is T, so the conjunction is F. A conjunction needs both properties of this same object. A successful P-witness at one row cannot be combined with a Q-witness at a different row as though the two names denoted one object.
exists x P(x): T. The first extension contains its declared object, so the first premise has a witness.
∃x Q(x): T. The second extension is also nonempty, so the second premise is true.
exists x (P(x) and Q(x)): F. The two extensions are disjoint.
Merge the two witness roles by putting the P-witness into Q as well as P. The conclusion becomes true, so that modified interpretation is no longer a countermodel. The original countermodel still establishes invalidity: one failed truth-preservation case suffices. This comparison distinguishes a countermodel from a mere model of the premises and prevents a common mistake of assuming that every interpretation must refute an invalid argument.
A small countermodel is often easier to audit than a complicated story. Begin by asking what must make the conclusion false. For an existential conjunction, arrange disjoint extensions so there is no shared witness. Then ask what must keep each premise true. Give each individual extension at least one member. Two distinct objects are enough to meet both demands.
Check whether a one-object domain could work. If the only object witnesses P and also witnesses Q, it must witness their conjunction, so the proposed conclusion cannot be false while both premises are true. This explains why the two-object construction is minimal. Minimality is useful for understanding the error, though an invalidity demonstration does not always need the smallest possible countermodel.
Other arguments fail for different reasons. From forall x (P(x) -> Q(x)) and exists x Q(x), existence of P does not follow. Set P empty and Q nonempty. The conditional universal is true because there is no P-counterexample, and the Q-existential has a witness. The P-existential conclusion is false. Here the issue is the direction of the conditional rather than a shared-witness mistake.
Record every extension, including an empty one. Leaving a predicate unspecified does not create a complete interpretation, so it may hide whether a premise is really true. An effective countermodel comes with a short audit: domain, extensions, premise values, conclusion value. Its explanatory power lies in showing how the premises leave room for the conclusion to fail. It challenges the inference's truth-preservation guarantee while leaving open whether the conclusion happens to hold in some intended real situation.
A countermodel satisfies premises while refuting a conclusion. An inconsistent set of premises instead has no model satisfying all of them. These are different diagnoses. If an interpretation fails a premise, it cannot serve as a countermodel to an inference from those premises, although it may help expose why a proposed specification cannot be satisfied.
Russell's paradox illustrates a restriction on an unrestricted collection-forming assumption. Suppose an object s is required to collect exactly the objects that do not belong to themselves. Write its membership requirement as R(s,x) iff not R(x,x), for every x. Instantiating at s yields R(s,s) iff not R(s,s). If the diagonal statement is true, the right side is false; if it is false, the right side is true. Neither classical value satisfies the required equivalence.
The failure does not show that classical inference rules permit an arbitrary contradiction from consistent premises. It shows that the unrestricted specification of this object cannot have the intended classical model. Standard axiomatic set theories restrict the collection-forming assumptions; they do not accept every grammatically describable collection as an unrestricted set. The course's relation-table audit isolates the diagonal conflict without pretending to teach an entire set theory.
Self-reference alone is not enough to produce a contradiction. An ordinary self-pair R(a,a) can be true or false consistently. The troublesome requirement links that pair to its own negation by a biconditional that is demanded to hold. Locate that exact requirement before calling a puzzling example a paradox. This also distinguishes the membership paradox from semantic puzzles about truth, which introduce different vocabulary and assumptions.
A community event needs somebody who both speaks the visitor's language and knows the building's accessible route. The organizer knows that one volunteer speaks the language and that another knows the route. It is tempting to conclude that a volunteer with both qualifications is available. The two existence facts do not establish that shared witness.
Build a two-person model. Ari speaks the language but does not know the route. Bo knows the route but does not speak the language. Both premise statements are true. The conclusion that somebody has both qualifications is false. The model uses invented roles rather than making claims about real volunteers, and it identifies exactly what information is missing.
The organizer might solve the practical problem by assigning the two volunteers together. That is a different plan with a different requirement. It does not turn the original one-person conclusion into a valid consequence of the two premises. Alternatively, a further check might find a third volunteer with both qualifications. This would make the conclusion true in the updated situation, without repairing the inference rule in general.
A useful countermodel therefore does more than announce wrong. It shows a possible arrangement of the stated facts in which the proposed conclusion fails. Write each person's properties explicitly and check each premise before the conclusion. If the description accidentally gives Ari both qualifications, it ceases to be a countermodel. The missing link is the intersection of the two qualifications, not the existence of each qualification somewhere in the group.
A correct evaluation answers the stated question for its stated interpretation. Do not turn a true instance into a universal rule or a successful example into a proof of validity. When the task is a proof audit, keep local assumptions and fresh parameters within their declared scope.
Record the interpretation and the question.
Domain: a, b, c, all distinct. P holds exactly of a; Q holds exactly of b. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
Use the declared objects and meanings throughout this calculation: exists x P(x) is the first requested result.
Determine the requested value: exists x P(x).
T
The first extension contains its declared object, so the first premise has a witness.
Determine the requested value: ∃x Q(x).
T
The second extension is also nonempty, so the second premise is true.
Determine the requested value: exists x (P(x) and Q(x)).
F
The two extensions are disjoint.
Collect the results in the requested order.
T / T / F
Each result belongs to its own entry: exists x P(x); ∃x Q(x); exists x (P(x) and Q(x)).
Record the interpretation and the question.
Domain: a, b, c, d, all distinct. P holds exactly of a; Q holds exactly of b. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
Use the declared objects and meanings throughout this calculation: exists x P(x) is the first requested result.
Determine the requested value: exists x P(x).
T
The first extension contains its declared object, so the first premise has a witness.
Determine the requested value: ∃x Q(x).
T
The second extension is also nonempty, so the second premise is true.
Determine the requested value: exists x (P(x) and Q(x)).
F
The two extensions are disjoint.
Collect the results in the requested order.
T / T / F
Each result belongs to its own entry: exists x P(x); ∃x Q(x); exists x (P(x) and Q(x)).
Record the interpretation and the question.
Domain: a, b, c, d, e, all distinct. P holds exactly of a; Q holds exactly of b. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
Use the declared objects and meanings throughout this calculation: exists x P(x) is the first requested result.
Determine the requested value: exists x P(x).
T
The first extension contains its declared object, so the first premise has a witness.
Determine the requested value: ∃x Q(x).
T
The second extension is also nonempty, so the second premise is true.
Determine the requested value: exists x (P(x) and Q(x)).
F
The two extensions are disjoint.
Collect the results in the requested order.
T / T / F
Each result belongs to its own entry: exists x P(x); ∃x Q(x); exists x (P(x) and Q(x)).
Test which alteration would change the conclusion.
Merge the two witness roles by putting the P-witness into Q as well as P. The conclusion becomes true, so that modified interpretation is no longer a countermodel. The original countermodel still establishes invalidity: one failed truth-preservation case suffices. This comparison distinguishes a countermodel from a mere model of the premises and prevents a common mistake of assuming that every interpretation must refute an invalid argument.
The altered interpretation checks the dependence of these answers on the stated model, rather than replacing it during the calculation.
Determine the requested value: exists x P(x).
T
The first extension contains its declared object, so the first premise has a witness.
Determine the requested value: ∃x Q(x).
Determine the requested value: exists x (P(x) and Q(x)).
Domain: a, b, c, d, e, all distinct. P holds exactly of c; Q holds exactly of d. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
| Computed result | |
|---|---|
| exists x P(x) | |
| ∃x Q(x) | |
| exists x (P(x) and Q(x)) |
Domain: a, b, c, d, e, f, g, all distinct. P holds exactly of e; Q holds exactly of f. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
Find a witness in the first extension.
b0
The requested entry concerns exists x p(x); retain its stated scope.
Find a witness in the second extension.
b1
The requested entry concerns ∃x q(x); retain its stated scope.
Inspect the intersection for a shared witness.
b2
The requested entry concerns exists x (p(x) and q(x)); retain its stated scope.
Domain: a, b, c, d, e, f, all distinct. P holds exactly of c; Q holds exactly of d. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
exists x P(x): b0
∃x Q(x): b1
exists x (P(x) and Q(x)): b2
Domain: a, b, c, d, e, all distinct. P holds exactly of e; Q holds exactly of a. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
exists x P(x): b0
∃x Q(x): b1
exists x (P(x) and Q(x)): b2
A proposed membership rule requires R(s,x) iff not R(x,x) for every object x, including s. At x=s, write the two-row truth audit. Columns are R(s,s), not R(s,s), and their biconditional. First row assumes R(s,s)=1, second assumes R(s,s)=0. Use 1 for true, 0 for false.
This task has no paper form; do it on a device.
For an event staffing audit, P means speaks the visitor's language and Q means knows the accessible route. The data below form the complete invented audit. Domain: a, b, c, d, e, f, all distinct. P holds exactly of a; Q holds exactly of b. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
exists x P(x): b0
∃x Q(x): b1
exists x (P(x) and Q(x)): b2
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Domain: a, b, c, d, e, f, g, all distinct. P holds exactly of g; Q holds exactly of a. Assess the inference from exists x P(x) and ∃x Q(x) to exists x (P(x) and Q(x)). Write the two premise truth values, then the conclusion truth value (T or F).
exists x P(x): b0
∃x Q(x): b1
exists x (P(x) and Q(x)): b2
You can supply a finite countermodel to a stated quantified inference. Reconstruct the three audit entries from a fresh model without consulting the examples; explain what change to the interpretation would change one answer.
15. Complete the next model audit, step 2
T
The second extension is also nonempty, so the second premise is true.
15. Complete the next model audit, step 3
F
The two extensions are disjoint.