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# notes_14_2x2 - Branden Fitelson Philosophy 12A Notes 1&...

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Unformatted text preview: Branden Fitelson Philosophy 12A Notes 1 ' & \$ % Announcements & Such • Robin Trower : Twice Removed from Yesterday • Administrative Stuff – HW #2 solutions posted. [Don’t sweat symbolization too much.] – I have posted two handouts: (1) solutions to problems from lecture on logical truth, equivalence, etc., and (2) three examples of the “short” truth-table method for validity (to be discussed today). – HW #3 has been posted. It’s on truth-table methods for validity. • Today: Chapter 3, Continued — More on LSL Semantics – The “short” method for testing LSL validity (continued). – An important connection between → and . – Some facts about semantic the semantic consequence relation ( ). – Expressive Completeness. UCB Philosophy Chapter 3 (Cont’d) 09/29/08 Branden Fitelson Philosophy 12A Notes 2 ' & The “Short” Method for Constructing Interpretations: Problem #1 • Question: A → (C ∨ E),B → D ? (A ∨ B) → (C → (D ∨ E)) . • Answer: A → (C ∨ E),B → D (A ∨ B) → (C → (D ∨ E)) . • Step 1: Assume there is an interpretation on which the premises is > but the conclusion is ⊥ . This leads to the following partial row: A B C D E A → (C ∨ E) B → D (A ∨ B) → (C → (D ∨ E)) > > ⊥ • Step 2: There’s only one way the conclusion can be ⊥ , which leads to: A B C D E A → (C ∨ E) B → D (A ∨ B) → (C → (D ∨ E)) > > > ⊥ ⊥ • Step 3: There’s only one way C → (D ∨ E) can be ⊥ , which leads to: A B C D E A → (C ∨ E) B → D (A ∨ B) → (C → (D ∨ E)) > > > > ⊥ > ⊥ ⊥ UCB Philosophy Chapter 3 (Cont’d) 09/29/08 Branden Fitelson Philosophy 12A Notes 3 ' & \$ % • Step 4: There’s only one way D ∨ E can be ⊥ , which leads to: A B C D E A → (C ∨ E) B → D (A ∨ B) → (C → (D ∨ E)) > ⊥ ⊥ > >⊥ > ⊥ > ⊥ ⊥ • Step 5: Since D is ⊥ , the only way B → D can be > is if B is ⊥ : A B C D E A → (C ∨ E) B → D (A ∨ B) → (C → (D ∨ E)) ⊥ > ⊥ ⊥ > ⊥>⊥ > ⊥ > ⊥ ⊥ • Step 6: Since C is > , so is C ∨ E , which makes A → (C ∨ E) > regardless of the truth-value of A . So, I will just let A be > , and then we’re done. A B C D E A → (C ∨ E) B → D (A ∨ B) → (C → (D ∨ E)) > ⊥ > ⊥ ⊥ >> > ⊥>⊥ > ⊥ > ⊥ ⊥ • When reporting your answer, all you need to do is give the single row that serves as a counterexample. Here, I recommend you include the quasi-columns that you used to calculate the truth-values in the row. • Verbal explanations are optional. Here’s the detailed handout solution. UCB Philosophy Chapter 3 (Cont’d) 09/29/08 Branden Fitelson Philosophy 12A Notes 4 ' & Answer . A → (C ∨ E),B → D (A ∨ B) → (C → (D ∨ E)) Explanation ....
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## This note was uploaded on 09/23/2009 for the course PHIL 12A taught by Professor Fitelson during the Spring '08 term at Berkeley.

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notes_14_2x2 - Branden Fitelson Philosophy 12A Notes 1&...

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