Give an example of a model (including domain, reference and extension) that has exactly two objects in its domain and on which all of the following four propositions are true.∃y∃x(x ≠ y ∧ Syxy) → ∀z(P z → Qz) ∀x∀y∀z(Sxyz ↔ (x = z ∧ y = a))∀x(Qx → x = b) a ≠ b ∧ (Qb ↔ ∃xSxbx)
Question
Give an example of a model (including domain, reference and extension) that has exactly two objects in its domain and on which all of the following four propositions are true.∃y∃x(x ≠ y ∧ Syxy) → ∀z(P z → Qz) ∀x∀y∀z(Sxyz ↔ (x = z ∧ y = a))∀x(Qx → x = b) a ≠ b ∧ (Qb ↔ ∃xSxbx)
Solution
The question seems to be asking for a model in the context of predicate logic. Here's an example:
Let's define our domain D = {a, b} where a and b are distinct objects.
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∃y∃x(x ≠ y ∧ Syxy) → ∀z(Pz → Qz) This proposition states that if there exists some y and x such that x is not equal to y and Syxy holds, then for all z, if Pz holds then Qz must also hold. In our model, we can say that Syxy is true only when x and y are different, which is always the case in our domain. Therefore, the antecedent of the implication is true. To make the whole implication true, we can define Pz and Qz to be false for all z in our domain. This way, the consequent of the implication is also true (since in logic, if the antecedent is false, the implication is true).
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∀x∀y∀z(Sxyz ↔ (x = z ∧ y = a)) This proposition states that for all x, y, and z, Sxyz is true if and only if x is equal to z and y is equal to a. In our model, we can define Sxyz to be true only when x = z and y = a. This makes the proposition true for all x, y, and z in our domain.
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∀x(Qx → x = b) This proposition states that for all x, if Qx is true then x must be equal to b. In our model, we can define Qx to be true only when x = b. This makes the proposition true for all x in our domain.
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a ≠ b ∧ (Qb ↔ ∃xSxbx) This proposition states that a is not equal to b and Qb is true if and only if there exists some x such that Sxbx is true. In our model, a and b are distinct, so the first part of the conjunction is true. We have defined Qb to be true and Sxbx to be true only when x = b, so the second part of the conjunction is also true.
Therefore, all four propositions are true in our model.
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