将指定谓词逻辑论证转为论证形式并运用推理规则证明其有效性

Let’s break down the problem into clear steps: first formalizing the argument using predicate logic, then proving its validity with standard inference rules.

Predicate Definitions #

We’ll use these predicates for all students in the class (our domain):

• P(x): x has taken Calculus I
• Q(x): x has taken Programming I
• R(x): x has instructor permission to take this course

Argument Form #

Premises #

1. Universal Disjunction: Every student has taken Calculus I or Programming I
   ∀x (P(x) ∨ Q(x))
2. Permission Conditional: Any student who took Programming I but not Calculus I has permission
   ∀x ((¬P(x) ∧ Q(x)) → R(x))

Conclusion to Derive #

The logical conclusion we’ll prove is:
All students without instructor permission have taken Calculus I
∀x (¬R(x) → P(x))

Proof of Validity (Using Inference Rules) #

We’ll use predicate and propositional logic rules to build the proof step-by-step, starting with an arbitrary student a (since universal statements apply to every member of the domain):

1. P(a) ∨ Q(a)
   Reason: Universal Instantiation (UI) on Premise 1—this rule applies to every student, so it holds for our arbitrary student a.
2. (¬P(a) ∧ Q(a)) → R(a)
   Reason: Universal Instantiation (UI) on Premise 2—same logic, the premise applies to all students including a.
3. Assume ¬R(a)
   Reason: We’re using Conditional Proof here—we want to show ¬R(a) → P(a), so we assume the antecedent (¬R(a)) and aim to derive the consequent (P(a)).
4. ¬(¬P(a) ∧ Q(a))
   Reason: Modus Tollens on steps 2 and 3. If A→B is true and ¬B is true, then ¬A must be true. Here, A = (¬P(a) ∧ Q(a)) and B = R(a).
5. P(a) ∨ ¬Q(a)
   Reason: De Morgan’s Law on step 4. The negation of a conjunction (¬(X ∧ Y)) equals the disjunction of negations (¬X ∨ ¬Y). Translating that gives us this statement.
6. P(a)
   Reason: Resolution (constructive dilemma) on steps 1 and 5. We have two disjunctions:
   • Step 1: P(a) ∨ Q(a) (either P(a) is true, or Q(a) is true)
   • Step 5: P(a) ∨ ¬Q(a) (either P(a) is true, or Q(a) is false)
     No matter if Q(a) is true or false, P(a) must be true to satisfy both statements.
7. ¬R(a) → P(a)
   Reason: Conditional Proof—since assuming ¬R(a) led us to P(a), the implication holds for our arbitrary student a.
8. ∀x (¬R(x) → P(x))
   Reason: Universal Generalization (UG)—since a was an arbitrary student in the domain, the implication applies to all students.

This completes the proof: the conclusion logically follows from the premises, so the argument is valid.

内容的提问来源于stack exchange，提问作者John Larkos