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formalize a property claim over a declared domain
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 formalize a property claim over a declared domain, 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. |
Propositional logic treats an atomic sentence as an indivisible letter. First-order logic opens some of that structure. A name denotes an object; a one-place predicate represents a property of objects. With P for being a participant and a for Amira, P(a) says that Amira is a participant. The choice of letters is a key, not evidence that the sentence is true.
An interpretation supplies a nonempty domain, a denotation for each name, and an extension for each predicate. The extension of a one-place predicate is a set of domain objects. P(a) is true exactly when the denotation of a belongs to that set. Distinct names need not denote distinct objects unless this is stated. Our early finite worksheets declare distinctness so that the membership calculation is unambiguous; the identity lesson later removes that convenience.
A variable such as x does not act like a fixed name. P(x) is an open formula until an assignment supplies a value for x or a quantifier binds it. An assignment can change while the interpretation stays fixed. This allows the same open formula to be checked at different objects. A closed sentence has no free variables left needing such an assignment. Keep this syntax question separate from whether the formula is true under a particular interpretation.
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.
The domain consists exactly of a, b; these names denote distinct objects. P is true exactly of a, b. The variable assignment gives x the object a. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
For object a, P(a) is T. The extension explicitly includes a, so this named object satisfies P. This is a membership fact about one object; it does not by itself say that another object satisfies P.
For object b, P(b) is T. The extension explicitly includes b, so this named object satisfies P. This is a membership fact about one object; it does not by itself say that another object satisfies P.
Objects in the extension: 2. Count objects in the extension of P, not characters in the predicate's name. The extension is the collection of domain objects for which the property is true. A predicate and its extension have different jobs: one belongs to the language, while the other belongs to this interpretation. A changed interpretation can give the same predicate symbol a different extension.
P(a): T. Interpret the name a first, then check that object's membership in P's extension. The expression P(a) is a closed atomic sentence because its argument is a name. It does not contain an unassigned variable. Its truth follows from this interpretation, not from the appearance of the letter P or an everyday meaning silently attached to it.
P(x) under the assignment: T. The assignment makes x refer to a for this evaluation. Substitute that object when checking the property, while leaving the predicate's extension fixed. P(x) is an open formula before an assignment or quantifier supplies the variable's value. An assignment is not an additional member of the domain and does not enlarge the set of objects.
Change the variable assignment while keeping the domain and the extension fixed. The truth of the open formula may change if the new object has different membership. The truth of P(a) will remain the same because neither the interpretation of a nor the extension of P has changed. This separates three things that a hurried reading can blur: the vocabulary, the interpretation, and the assignment.
The domain consists exactly of a, b, c; these names denote distinct objects. P is true exactly of a. The variable assignment gives x the object b. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
For object a, P(a) is T. The extension explicitly includes a, so this named object satisfies P. This is a membership fact about one object; it does not by itself say that another object satisfies P.
For object b, P(b) is F. The complete extension omits b, so this named object does not satisfy P. Omission here means false because the interpretation is complete; an incomplete real-world record would not license the same assumption.
For object c, P(c) is F. The complete extension omits c, so this named object does not satisfy P. Omission here means false because the interpretation is complete; an incomplete real-world record would not license the same assumption.
Objects in the extension: 1. Count objects in the extension of P, not characters in the predicate's name.
P(a): T. Interpret the name a first, then check that object's membership in P's extension.
P(x) under the assignment: F. The assignment makes x refer to b for this evaluation.
Change the variable assignment while keeping the domain and the extension fixed. The truth of the open formula may change if the new object has different membership. The truth of P(a) will remain the same because neither the interpretation of a nor the extension of P has changed. This separates three things that a hurried reading can blur: the vocabulary, the interpretation, and the assignment.
The domain consists exactly of a, b, c, d; these names denote distinct objects. P is true exactly of a, b, c. The variable assignment gives x the object c. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
For object a, P(a) is T. The extension explicitly includes a, so this named object satisfies P. This is a membership fact about one object; it does not by itself say that another object satisfies P.
For object b, P(b) is T. The extension explicitly includes b, so this named object satisfies P. This is a membership fact about one object; it does not by itself say that another object satisfies P.
For object c, P(c) is T. The extension explicitly includes c, so this named object satisfies P. This is a membership fact about one object; it does not by itself say that another object satisfies P.
For object d, P(d) is F. The complete extension omits d, so this named object does not satisfy P. Omission here means false because the interpretation is complete; an incomplete real-world record would not license the same assumption.
Objects in the extension: 3. Count objects in the extension of P, not characters in the predicate's name.
P(a): T. Interpret the name a first, then check that object's membership in P's extension.
P(x) under the assignment: T. The assignment makes x refer to c for this evaluation.
Change the variable assignment while keeping the domain and the extension fixed. The truth of the open formula may change if the new object has different membership. The truth of P(a) will remain the same because neither the interpretation of a nor the extension of P has changed. This separates three things that a hurried reading can blur: the vocabulary, the interpretation, and the assignment.
Consider two dictionaries for the same situation. The first uses P for wearing a pass and a for Ada. The second uses W for wearing a pass and d for Ada. If the dictionaries explicitly give these meanings, P(a) in the first language and W(d) in the second describe the same object having the same property. A difference in printed letters need not be a difference in meaning.
Now keep the first dictionary but change P's extension from Ada alone to Ada and Bo. That is a genuine change of interpretation. P(a) remains true, yet the number of P-objects changes. The unchanged truth of one sentence does not show that the models are identical. Other sentences, including an existential statement about somebody other than Ada, can distinguish them.
Arity also belongs to the dictionary. A one-place predicate expects one argument, while a two-place predicate expects two. If B(x,y) means x borrows from y, B(a) is not merely a false sentence; it has failed to supply the required argument. Distinguish malformed syntax from a well-formed false statement. A model evaluates formulas after their syntax has been fixed.
The same discipline helps with free variables. P(x) and P(y) can express corresponding open conditions under suitably renamed assignments, but changing a name into a variable is not just renaming. The variable's value may depend on an assignment or quantifier. Before comparing two expressions, ask which differences concern notation, which concern the interpretation, and which concern the assignment. This three-part check prevents an apparently tiny edit from silently changing the task being solved.
Imagine a workshop with exactly four registered participants: Ada, Bo, Cy and Di. Let P mean has collected a materials pack. The organizer's checked register says that Ada and Cy have collected theirs; Bo and Di have not. The domain is the four registered people, and P's extension is the two-person set containing Ada and Cy. A name for Ada makes P of that name true. A name for Bo makes the corresponding sentence false.
This formalization helps separate the people from the property being recorded. Changing the domain to all visitors would require including visitors who are not registered. Changing P to has paid would require a different extension, even if some of the same people belonged to it. Neither change is just a harmless change of title: it changes what the model says.
Now suppose the organizer only has an old note listing Ada. Failure to find Bo on that incomplete note would not establish that Bo lacks a pack. The finite exercises use complete interpretations: unlisted members are explicitly outside the extension. A real record needs its completeness checked before the same inference is used. Write down whether a blank means no, not yet checked, or missing data.
Finally, assign x to Cy and inspect P(x). It is true under that assignment. Assign x to Di and it is false, although the register itself has not changed. This is the practical distinction between selecting a record to inspect and changing the underlying data. A precise interpretation makes each of these operations visible.
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.
The domain consists exactly of a, b, c; these names denote distinct objects. P is true exactly of a, b. The variable assignment gives x the object a. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
Use the declared objects and meanings throughout this calculation: Objects in the extension is the first requested result.
Determine the requested value: Objects in the extension.
2
Count objects in the extension of P, not characters in the predicate's name.
Determine the requested value: P(a).
T
Interpret the name a first, then check that object's membership in P's extension.
Determine the requested value: P(x) under the assignment.
T
The assignment makes x refer to a for this evaluation.
Collect the results in the requested order.
2 / T / T
Each result belongs to its own entry: Objects in the extension; P(a); P(x) under the assignment.
Record the interpretation and the question.
The domain consists exactly of a, b, c, d; these names denote distinct objects. P is true exactly of a, b, c, d. The variable assignment gives x the object a. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
Use the declared objects and meanings throughout this calculation: Objects in the extension is the first requested result.
Determine the requested value: Objects in the extension.
4
Count objects in the extension of P, not characters in the predicate's name.
Determine the requested value: P(a).
T
Interpret the name a first, then check that object's membership in P's extension.
Determine the requested value: P(x) under the assignment.
T
The assignment makes x refer to a for this evaluation.
Collect the results in the requested order.
4 / T / T
Each result belongs to its own entry: Objects in the extension; P(a); P(x) under the assignment.
Record the interpretation and the question.
The domain consists exactly of a, b, c, d, e; these names denote distinct objects. P is true exactly of none. The variable assignment gives x the object a. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
Use the declared objects and meanings throughout this calculation: Objects in the extension is the first requested result.
Determine the requested value: Objects in the extension.
0
Count objects in the extension of P, not characters in the predicate's name.
Determine the requested value: P(a).
F
Interpret the name a first, then check that object's membership in P's extension.
Determine the requested value: P(x) under the assignment.
F
The assignment makes x refer to a for this evaluation.
Collect the results in the requested order.
0 / F / F
Each result belongs to its own entry: Objects in the extension; P(a); P(x) under the assignment.
Test which alteration would change the conclusion.
Change the variable assignment while keeping the domain and the extension fixed. The truth of the open formula may change if the new object has different membership. The truth of P(a) will remain the same because neither the interpretation of a nor the extension of P has changed. This separates three things that a hurried reading can blur: the vocabulary, the interpretation, and the assignment.
The altered interpretation checks the dependence of these answers on the stated model, rather than replacing it during the calculation.
Determine the requested value: Objects in the extension.
3
Count objects in the extension of P, not characters in the predicate's name.
Determine the requested value: P(a).
Determine the requested value: P(x) under the assignment.
The domain consists exactly of a, b, c, d, e; these names denote distinct objects. P is true exactly of a, b, c, d. The variable assignment gives x the object c. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
| Computed result | |
|---|---|
| Objects in the extension | |
| P(a) | |
| P(x) under the assignment |
The domain consists exactly of a, b, c, d, e, f, g; these names denote distinct objects. P is true exactly of a, b, c, d, e, f. The variable assignment gives x the object e. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
Count the extension's objects.
b0
The requested entry concerns objects in the extension; retain its stated scope.
Resolve the named argument before testing membership.
b1
The requested entry concerns p(a); retain its stated scope.
Use the assigned value of the free variable.
b2
The requested entry concerns p(x) under the assignment; retain its stated scope.
The domain consists exactly of a, b, c, d, e, f; these names denote distinct objects. P is true exactly of a, b, c, d, e, f. The variable assignment gives x the object c. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
Objects in the extension: b0
P(a): b1
P(x) under the assignment: b2
The domain consists exactly of a, b, c, d, e; these names denote distinct objects. P is true exactly of a, b, c, d. The variable assignment gives x the object e. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
Objects in the extension: b0
P(a): b1
P(x) under the assignment: b2
The domain consists exactly of a, b, c, d, e, f; these names denote distinct objects. P is true exactly of a, b, c, d, e. The variable assignment gives x the object e. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
Objects in the extension: b0
P(a): b1
P(x) under the assignment: b2
In a workshop register, the domain objects are participants and P means has collected a pack. The data below form the complete invented audit. The domain consists exactly of a, b, c, d, e, f; these names denote distinct objects. P is true exactly of a, b, c, d, e, f. The variable assignment gives x the object a. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
Objects in the extension: b0
P(a): b1
P(x) under the assignment: b2
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The domain consists exactly of a, b, c, d, e, f, g; these names denote distinct objects. P is true exactly of a, b, c, d, e, f. The variable assignment gives x the object g. Write the number of objects satisfying P, the truth value of P(a), and the truth value of P(x) under that assignment. Use T or F for truth values.
Objects in the extension: b0
P(a): b1
P(x) under the assignment: b2
You can formalize a property claim over a declared domain. Reconstruct the three audit entries from a fresh model without consulting the examples; explain what change to the interpretation would change one answer.
14. Complete the next model audit, step 2
T
Interpret the name a first, then check that object's membership in P's extension.
14. Complete the next model audit, step 3
T
The assignment makes x refer to c for this evaluation.