1 / 8

INTRO LOGIC

This document outlines the main content for Exam 4 on Predicate Logic, highlighting key derivations, including universal and existential quantifiers, negation rules, and polyadic quantifiers. Each derivation is structured with example exercises and points distribution, encouraging a comprehensive grasp of the subject. The exam covers multiple quantifier interactions and how they can be manipulated through logical derivations. Students are advised to review exercises from Sets C, D, E, F, and G, and best of luck on the exams!

holden
Télécharger la présentation

INTRO LOGIC

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. INTRO LOGIC Derivations in PL5 DAY 26

  2. Exam 4 Topics • 6 derivations in Predicate Logic • 15 points each, plus 10 free points • universal derivation [Exercise Set C] • existential-out [Exercise Set D] • negation rules [Exercise Set E] • multiple quantifiers [Exercise Set F] • polyadic quantifiers [Exercise Set G] • polyadic quantifiers [Exercise Set G]

  3. Example 1 (repeated from last lecture) there is someone whom everyone R’s/ everyone R’s someone or other (1) xyRyx Pr (2) : xyRxy UD (3) : yRay D(ID) (4) yRay As (5) :  DD (6) yRyb 1, O (7) yRay 4, O (8) Rab 6, O (9) Rab 7, O (10)  8,9, I

  4. Example 2 • there is a F who R’s every G/ every G is R’ed by some F or other (1) x(Fx & y(Gy  Rxy)) Pr (2) : x(Gx y(Fy & Ryx))

  5. (1) x(Fx & y(Gy  Rxy)) Pr (2) : x(Gx y(Fy & Ryx)) UD (3) : Gay(Fy & Rya) CD (4) Ga As (5) : y(Fy & Rya) D(ID) (6) y(Fy & Rya) As (7) :  DD (9) Fb & y(Gy  Rby) 1, O (8) y(Fy & Rya) 6, O (10) Fb 9, &O (11) y(Gy  Rby) (12) (Fb & Rba) 8, O (13) Ga  Rba 11, O (14) Fb Rba 12, &O (15) Rba 4,13, O (16) Rba 10,14 O (17)  15,16, I

  6. Example 3 • there is a G who R’s no F/ every F is dis-R’ed by at least one G (1) x(Gx & y(Fy & Rxy)) Pr (2) : x(Fx y(Gy & Ryx))

  7. (1) x(Gx & y(Fy & Rxy)) Pr (2) : x(Fx y(Gy & Ryx)) UD (3) : Fay(Gy & Rya) CD (4) Fa As (5) : y(Gy & Rya) D (ID) (6) y(Gy & Rya) As (7) :  DD (8) Gb & y(Fy & Rby) 1, O (9) y(Gy & Rya) 6, O (10) Gb 8, &O (11) y(Fy & Rby) (12) (Gb & Rba) 9, O (13) y(Fy & Rby) 11, O (14) Gb Rba 12, &O (15) (Fa & Rba) 13, O (16) Fa Rba 15, &O (17) Rba 10,1, O (18) Rba 4,16, O (19)  17,18, I

  8. GOOD LUCK ON EXAM 4! GOOD LUCK ON THE FINALS! HAVE A GREAT SUMMER! GO SOX!

More Related