
Liam B. answered 12/20/20
Computer Science Student with Teaching and Industry Experience
The first three questions require translation, while the last requires a logical proof.
a) ∀x(P(x) ⊃ Q(x)) When you see "All" it's going to be a universal quantifier.
b) ∃x(R(x) ∧ ¬Q(x)) Likewise when you see "Some" it's an existential quantifier.
c) ∃x(R(x) ∧ ¬P(x))
Another "trick" is that for statements like these, universal quantifiers usually have a conditional inside them, whereas existential quantifiers will have a conjunction. This isn't a hard and fast rule, but it's a good assumption.
d) Yes
- R(α) ∧ ¬Q(α), b Existential instantiation. α is a free variable (Note: To be a neat proof this step should come before 2. The reason is because it's an existential rather than universal instantiation. i.e α might be the only x that is both R and not Q)
- P(α) ⊃ Q(α), a Universal instantiation
- ¬Q(α), 1 Simplification
- ¬P(α), 2 & 4 Modus tollens
- R(α), 1 Simplification
- R(α) ∧ ¬P(α), 4 & 5 Conjunction
∴ ∃x(R(x) ∧ ¬P(x)), 6 Existential generalization