Lecture #12 - Branden Fitelson Philosophy 12A Notes 1& $...

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Unformatted text preview: Branden Fitelson Philosophy 12A Notes 1 ' & $ % Announcements & Such • Dire Straits : In the Gallery • Administrative Stuff – HW #2 solutions posted. [Don’t sweat symbolizations too much.] + Hints to controversial problems were on my “tips” handout. – Two handouts: (1) solutions to problems from lecture on logical truth, equivalence, etc.; (2) 3 “short” truth-table method problems. + Make sure you study my handouts. They tend to be useful. – HW #3 due Today , usual drill (truth-table methods for validity). – Branden won’t hold office hours today [there’s a colloquium]. • Today: Chapter 3, Finalé — Some Final Topics on LSL Semantics – Expressive Completeness: re-cap + some additional remarks. – Some Chapter 2 Problems — in Light of Chapter 3 – Knights & Knaves Puzzles, and an LSAT Problem • Next: Chapter 4 — Natural Deduction Proofs in LSL UCB Philosophy Chapter 3 Finalé 02/25/10 Branden Fitelson Philosophy 12A Notes 2 ' & $ % Expressive Completeness: Recap • Fact . The set of 4 connectives h∼ , & , ∨ , →i is expressively complete. [ p ↔ q , [ (p → q) & (q → p) • Fact . The set of 3 connectives h∼ , & , ∨i is expressively complete. [ p → q , [ ∼ p ∨ q • Fact . The pairs h∼ , & i and h∼ , ∨i are both expressively complete. [ p ∨ q , [ ∼ ( ∼ p & ∼ q) – The h∼ , ∨i strategy is similar [ [ p & q , [ ∼ ( ∼ p ∨∼ q) ]. • Consider the binary connective ‘ | ’ such that [ p | q [ ∼ (p & q) . • Fact . ‘ | ’ alone is expressively complete! How to express h∼ , & i using ‘ | ’: [ ∼ p , [ p | p , and [ p & q , [ (p | q) | (p | q) – I called ‘ | ’ ‘NAND’ in a previous lecture. NOR is also expressively complete. UCB Philosophy Chapter 3 Finalé 02/25/10 Branden Fitelson Philosophy 12A Notes 3 $ % Expressive Completeness: Additional Remarks and Questions • Q . How can we define ↔ in terms of | ? A . If you naïvely apply the schemes I described last time, then you get a 187 symbol monster : [ p ↔ q , A | A , where A is given by the following 93 symbol expression: (((p | (q | q)) | (p | (q | q))) | ((p | (q | q)) | (p | (q | q)))) | (((q | (p | p)) | (q | (p | p))) | ((q | (p | p)) | (q | (p | p)))) • There are simpler definitions of ↔ using | . E.g. , this 43 symbol answer: [ p ↔ q , ((p | (q | q)) | (q | (p | p))) | ((p | (q | q)) | (q | (p | p))) • Can anyone give an even simpler definition of ↔ using | ? Extra-Credit! • How could you show that the pair h→ , ∼i is expressively complete? • Fact . No subset of h∼ , & , ∨ , → , ↔i that does not contain negation ∼ is expressively complete. [This is a 140A question, beyond our scope.] • Let denote the ⊥ truth-function ( i.e. , the trivial function that always returns ⊥ ). How could you show that h→ , i is expressively complete?...
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Lecture #12 - Branden Fitelson Philosophy 12A Notes 1& $...

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