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.
Question
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.
Solution
The given model is a bit complex, but let's break it down step by step.
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∃x¬(x = a → Lxx): This proposition states that there exists an x such that it's not the case that if x equals a, then Lxx. In the given model, a refers to 1 and the extension of L is empty. Therefore, there is no x such that Lxx, so this proposition is true.
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∀y(¬Lay ∧ Ryb): This proposition states that for all y, it's not the case that Lay and Ryb. In the given model, a refers to 1, b refers to 2, and the extension of L is empty while the extension of R includes {1,2}. Therefore, it's true that for all y, ¬Lay. However, Ryb is only true for y=1 and y=2. Therefore, this proposition is true.
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¬Laa ∧ Rab: This proposition states that it's not the case that Laa and Rab. In the given model, a refers to 1, b refers to 2, and the extension of L is empty while the extension of R includes {1,2}. Therefore, ¬Laa is true and Rab is true. Therefore, this proposition is true.
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¬Lab ∧ Rbb: This proposition states that it's not the case that Lab and Rbb. In the given model, a refers to 1, b refers to 2, and the extension of L is empty while the extension of R includes {1,2} and {2,2}. Therefore, ¬Lab is true and Rbb is true. Therefore, this proposition is true.
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¬LaaRab¬LabRbb¬(c = a → Lcc): This proposition is a bit complex, but it essentially states that it's not the case that Laa, Rab, Lab, Rbb, and if c equals a, then Lcc. In the given model, a refers to 1, b refers to 2, c refers to 1, and the extension of L is empty while the extension of R includes {1,2} and {2,2}. Therefore, all the negations are true and since c equals a, ¬Lcc is also true. Therefore, this proposition is true.
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c = a ¬Lcc¬Lac ∧ Rcb: This proposition states that c equals a, it's not the case that Lcc, ¬Lac, and Rcb. In the given model, a refers to 1, b refers to 2, c refers to 1, and the extension of L is empty while the extension of R includes {1,2}. Therefore, c does equal a, ¬Lcc is true, ¬Lac is true, and Rcb is true. Therefore, this proposition is true.
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¬LacRcb¬Lca¬Lcb: This proposition states that it's not the case that Lac, Rcb, Lca, and Lcb. In the given model, a refers to 1, b refers to 2, c refers to 1, and the extension of L is empty while the extension of R includes {1,2}. Therefore, all the negations are true and Rcb is true. Therefore, this proposition is true.
Therefore, all of the propositions on the open path are true on the given model.
Similar Questions
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