Exam #1 is Friday. It will take the entire period. You do not need to bring an exam booklet; you’ll write directly on the exam. Outline: I. T/F, covering.

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Presentation transcript:

Exam #1 is Friday. It will take the entire period. You do not need to bring an exam booklet; you’ll write directly on the exam. Outline: I. T/F, covering validity and soundness (see problems at the power of logic 1.1B fixed.cgi?exercise=1.1B ), 7, worth 2 points each. fixed.cgi?exercise=1.1B fixed.cgi?exercise=1.1B

Outline of Exam #1, continued II. Symbolizing statements (full dictionary provided), five problems, worth 3, 4, 4, 5, and 5 pts. III. Symbolize an entire argument, worth 7 points IV. Two parse trees, worth 4 pts. each.

Symbolizing Complete Arguments An argument: After I get my B.A., I’ll either go to law school, medical school, or business school. I hear that law school is incredibly boring, and med school takes forever. An MBA it is, then. First determine the conclusion: B: I’m getting an MBA.

Other Atomic Statements? L: I go to law school after B.A. M: I go to med school after B.A. I: Law school is incredibly boring. F: Med school takes forever. S: I go to business school U: I get my B.A. If you don’t consider what the speaker/author means, this doesn’t give you a very sensible solution.

Not-so-good solution U & [L v (M v S)] I & F  B What’s a better solution? Think about the point of the argument.

For the purpose of this argument, going to business school is equivalent to getting an MBA. So, change S to B. Also, I and F are relevant because they’re equivalent to (or at least are meant to support) ~ L and ~ M. So, here’s a somewhat bare-bones, but correct, version. U & [L v (M v B)] ~ L & ~ M  B

A more complete version? U & [L v (M v B)] I & F I → ~ L F → ~ M  B

Formal Description of the Language of Sentential Logic (LSL) The lexicon: All capital letters (augmented by any number of “prime” marks); the five connectives; and parentheses.

Formation Rules for LSL Base clause (f-sl): Every sentence-letter of LSL is a well- formed formula (a wff) Recursion clauses: (f-~): If p is wff then the result of prefixing ‘~’ to p is a wff. (f-&): If p and q are wffs, then placing ‘&’ between them and enclosing the result in parentheses yields a wff. (f-v): If p and q are wffs, then placing ‘v’ between them and enclosing the result in parentheses yields a wff. (f-→): If p and q are wffs, then placing ‘→’ between them and enclosing the result in parentheses yields a wff. (f-↔): If p and q are wffs, then placing ‘↔’ between them and enclosing the result in parentheses yields a wff. Closure Clause: Nothing is a wff in LSL if it is not certified by the preceding rules.

Formation rules and parse trees For any wff p, you can construct a parse tree starting with capital letters at the bottom and adding nodes using only the recursion rules until you reach the top, which has a single node p. Any symbol string constructed in this way is a wff. And nothing that can’t be constructed in this way is a wff.

Subformulae ‘Formulae’ is the traditional plural form of ‘formula’ A subformula of p is any formula appearing at a node on p’s parse tree.

Scope and main connectives The scope of a binary connective c is the subformula appearing at the node where c is introduced. A tilde’s scope includes all and only the symbols appearing in the node immediately below the node introducing the tilde. The main connective of a formula is the last connective added to it, working upwards through the parse tree.

Exercises, p. 40