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Unformatted text preview: Introduction to Logic Lecture 11 Brian Weatherson, Department of Philosophy October 7, 2009 Logic 201 (Section 5) Lecture 11 1 Rule Revision Logic 201 (Section 5) Lecture 11 2 ∧Elimination From A ∧ B infer A . We cite the line where A ∧ B occurs. Logic 201 (Section 5) Lecture 11 3 ∧Elimination Also, from A ∧ B we can infer B . The same line is cited. Logic 201 (Section 5) Lecture 11 4 ∧Introduction From A and B , infer A ∧ B . We cite the lines where A and B occur. Logic 201 (Section 5) Lecture 11 5 ∨Introduction From A , infer A ∨ B . We cite the line where A occurs. Logic 201 (Section 5) Lecture 11 6 ∨Introduction Like with ∧elimination, there are two versions of this rule. You can also use it to get from B , to A ∨ B . In that case, we cite the line where B occurs. Logic 201 (Section 5) Lecture 11 7 ∨elimination When using ∨elimination, we need, and cite, three things. 1 The original disjunction. 2 The subproof from A to C . 3 The subproof from B to C . Logic 201 (Section 5) Lecture 11 8 ¬elimination From ¬¬ A infer A . Logic 201 (Section 5) Lecture 11 9 ⊥introduction From A and ¬ A infer ⊥ . In general, whenever you have two contradictory sentences, you can infer ⊥ . Logic 201 (Section 5) Lecture 11 10 ¬introduction From subproof that starts with A , ends with ⊥ , infer ¬ A . In general we’ll prove negated sentences this way. Note again that we cite a proof (see the hyphen in the justification of line 4). Logic 201 (Section 5) Lecture 11 11 ⊥elimination So ⊥elimination rule says, from ⊥ infer whatever you like. This will help shorten some otherwise tricky proofs we’ll later see. Logic 201 (Section 5) Lecture 11 12 Examples Logic 201 (Section 5) Lecture 11 13 Strategy We’ll go over two of the examples from last time, spending a bit more time on how the proofs were put together. We’ll start with A ∨ B C ( A ∧ C ) ∨ B Logic 201 (Section 5) Lecture 11 14 A ∨ B , C , therefore ( A ∧ C ) ∨ B 1 A ∨ B 2 C 3 A 4 A ∧ C ∧intro: 2, 3 5 ( A ∧ C ) ∨ B ∨intro: 4 6 B 7 ( A ∧ C ) ∨ B ∨intro: 6 8 ( A ∧ C ) ∨ B ∨elim: 1, 35, 67 We start with the disjunction in premise one, then show that the conclusion follows from each disjunct. That sets up a use of ∨elimination in the final line. Logic 201 (Section 5) Lecture 11 15 A ∨ B , C , therefore ( A ∧ C ) ∨ B If we were putting this together, the first thing we’d do is notice that we have a disjunctive premise, so we will eventually have to use ∨elimination. Let’s set that up: 1 A ∨ B 2 C 3 A 4 Rule? 5 ( A ∧ C ) ∨ B Rule? 6 B 7 ( A ∧ C ) ∨ B Rule? 8 ( A ∧ C ) ∨ B ∨elim: 1, 35, 67 Logic 201 (Section 5) Lecture 11 16 A ∨ B , C , therefore ( A ∧ C ) ∨ B The path from A to ( A ∧ C ) ∨ B goes via A ∧ C ....
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This note was uploaded on 02/01/2012 for the course 730 201 taught by Professor Jonwinterbottom during the Fall '11 term at Rutgers.
 Fall '11
 JonWinterbottom

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