# short - The “Short” Truth-Table Method Three Examples...

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Unformatted text preview: The “Short” Truth-Table Method: Three Examples Philosophy 12A June 3, 2010 1 Example #1 — Page 66 #3 Answer . A → (C ∨ E),B → D ø (A ∨ B) → (C → (D ∨ E)) Explanation . 1 Assume that ‘ A → (C ∨ E) ’ is > , ‘ B → D ’ is > , and ‘ (A ∨ B) → (C → (D ∨ E)) ’ is ⊥ . In order for ‘ (A ∨ B) → (C → (D ∨ E)) ’ to be ⊥ , both ‘ A ∨ B ’ and ‘ C ’ must be > , and both ‘ D ’ and ‘ E ’ must be ⊥ . This guarantees that the first premise is > (since ‘ A → (C ∨ E) ’ must , at this point, have a > consequent). We can also make the second premise > , simply by making ‘ B ’ ⊥ . Finally, by making ‘ A ’ > , we can ensure that the conclusion is ⊥ , which yields the following interpretation on which ‘ A → (C ∨ E) ’ and ‘ B → D ’ are > , but ‘ (A ∨ B) → (C → (D ∨ E)) ’ is ⊥ ( i.e. , the following counterexample to validity). A B C D E A → (C ∨ E) B → D (A ∨ B) → (C → (D ∨ E)) > ⊥ > ⊥ ⊥ > > > > ⊥ ⊥ > ⊥ > > ⊥ ⊥ > ⊥ ⊥ ⊥ ⊥ Therefore, by the definition of...
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## This note was uploaded on 11/26/2011 for the course PHILOSOPHY 101 taught by Professor Buechner during the Fall '06 term at Rutgers.

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