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# notes_23_2x2 - Branden Fitelson Philosophy 12A Notes 1&...

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Unformatted text preview: Branden Fitelson Philosophy 12A Notes 1 ' & \$ % Announcements & Such • Godspeed You Black Emperor : The Dead Flag Blues • Administrative Stuff – The in-class mid-term is on Friday. * Same structure as the sample mid-term. * Rules Handout will be provided at the exam (bring blue book). – The Take-Home Mid-Term resubmission is due on Friday . – I have posted the solutions for HW #3. – I have posted HW #4 (due next Friday). – Branden will not have office hours on Friday. • Today: Chapter 4 — Natural Deduction Proofs for LSL – The ∨ rules, and lots of (solved) proof problems. – Next: Sequent and Theorem Introduction (derived rules). • More on MacLogic for constructing and checking proofs. UCB Philosophy Chapter 4 10/22/08 Branden Fitelson Philosophy 12A Notes 2 ' & The Disjunction Rules Rule of ∨-Introduction : For any formula p , if p has been inferred at line j, then, for any formula q , either [ p ∨ q or [ q ∨ p may be inferred at line k, labeling the line ‘j ∨ I’ and writing on its left the same premise and assumption numbers as appear on the left of j. a 1 ,. . . , a n (j) p a 1 ,. . . , a n (j) q . . . OR . . . a 1 ,. . . , a n (k) p ∨ q j ∨ I a 1 ,. . . , a n (k) p ∨ q j ∨ I • The ∨ I rule is very simple an intuitive. Basically, it says that you may infer a disjunction from either of its disjuncts. • The elimination rule ( ∨ E) for ∨ , on the other hand, is considerably more complex to state and apply. It’s the hardest of our rules. UCB Philosophy Chapter 4 10/22/08 Branden Fitelson Philosophy 12A Notes 3 ' & \$ % Rule of ∨-Elimination : If a disjunction [ p ∨ q occurs at line g of a proof, p is assumed at line h, r is derived at line i, q is assumed at line j, and r is derived at line k, then at line m we may infer r , labeling the line ‘g, h, i, j, k ∨ E’ and writing on its left every number on the left at line g, and at line i (except h), and at line k (except j). a 1 ,. . . , a n (g) p ∨ q . . . h (h) p Assumption . . . b 1 ,. . . , b u (i) r . . . j (j) q Assumption . . . c 1 ,. . . , c w (k) r . . . A (m) r g, h, i, j, k ∨ E where A is the set: {a 1 ,. . . , a n } ∪ {b 1 ,. . . , b u }/h ∪ {c 1 ,. . . , c w }/j. UCB Philosophy Chapter 4 10/22/08 Branden Fitelson Philosophy 12A Notes 4 ' & Another Example Involving ∨ I and Negation • Here’s a proof of the theorem : ` A ∨ ∼ A . Problem is: Ê AÎÒA 1 (1) Ò(AÎÒA)nobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspaceAssumption (ÒI) 2 (2) A„„„nobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspaceAssumption (ÒI) 2 (3) AÎÒA„„nobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspace2 ÎI 1,2 (4) Ï„nobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspacenobreakspace1,3...
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notes_23_2x2 - Branden Fitelson Philosophy 12A Notes 1&...

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