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Context: Here is a model: Domain: {1, 2, 3, 4} B : {3, 4}, C : {1, 4}, F : ∅ Make a proposition that is false in this model by filling in each of the blank spaces of (blank)z((B(blank)(blank) z) →(blank)(blank)y¬F y) with one symbol of MPL. Explain why the proposition you made is false in the given model. Answer question

Question

Context: Here is a model: Domain: {1, 2, 3, 4} B : {3, 4}, C : {1, 4}, F : ∅ Make a proposition that is false in this model by filling in each of the blank spaces of (blank)z((B(blank)(blank) z) →(blank)(blank)y¬F y) with one symbol of MPL. Explain why the proposition you made is false in the given model. Answer question

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Solution

Sure, let's fill in the blanks with the symbols of MPL (Modal Propositional Logic) to create a proposition that is false in the given model.

Let's fill in the blanks as follows:

∀z((Bz ∧ ∃z) → ∃y¬Fy)

This proposition can be read as: "For all z, if z is in B and there exists such a z, then there exists a y such that y is not in F."

Now, let's see why this proposition is false in the given model.

In the given model, B includes {3, 4} and F is an empty set. According to the proposition, if a number is in B, then there should exist a number that is not in F. However, since F is an empty set, every number is not in F. Therefore, the proposition should be true for all numbers in B, which are 3 and 4.

However, the proposition is a universal statement, meaning it should hold true for all elements in the domain. The domain in the given model is {1, 2, 3, 4}. For the numbers 1 and 2, which are not in B, the proposition does not hold true because the implication "if z is in B then there exists a y such that y is not in F" is not satisfied. Therefore, the proposition is false in the given model.

This problem has been solved

Similar Questions

2. Here is a model: Domain: {1, 2, 3, 4} B : {3, 4}, C : {1, 4}, F : ∅ Make a proposition that is false in this model by filling in each of the blank spaces of _z((Bz _ _ z) → _ ∀ y¬F y) with one symbol of MPL. Explain why the proposition you made is false in the given model.

Consider the following saturated open path and model.∃x¬(x = a → Lxx) X c ∀y(¬Lay ∧ Ryb) \abc¬Laa ∧ Rab X¬Lab ∧ Rbb X¬LaaRab¬LabRbb¬(c = a → Lcc) X c = a ¬Lcc¬Lac ∧ Rcb X¬LacRcb¬Lca¬Lcb↑Domain: {1, 2, 3}Referents: a : 1 b : 2 c : 1Extension of R : {h1, 2i ,h 2, 2i ,h 1, 2i}Extension of L : ∅Are all of the propositions on the open path true on the given model? Explain your answer.

Give an example of a model 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)

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)

Consider the following two propositions.∀x∀y∀z((x 6 = y ∧ y 6 = z ∧ x 6 = z) → (¬P x ∨ ¬P y ∨ ¬P z))∃z(a = a → ∀yP y)(i) What is the maximum number of objects a model can have in its domain if both of the propositions are true on that model? Explain your answer.(ii) Give an example of a model that has the maximum number of objects in its domain and on which both of the propositions are true.

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