# existential elimination

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Existential Elimination. . Kareem Khalifa Department of Philosophy Middlebury College. Overview. An example Existential Elimination: 3 Steps Qualifications and tricks Examples. An example. Somebody in this class is a musician and a soccer player. Therefore, someone is a musician. - PowerPoint PPT Presentation

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• Existential EliminationKareem KhalifaDepartment of PhilosophyMiddlebury College

• OverviewAn exampleExistential Elimination: 3 StepsQualifications and tricksExamples

• An exampleSomebody in this class is a musician and a soccer player. Therefore, someone is a musician.x(Cx&(Mx&Sx)) xMxThis is clearly valid, yet we dont have a way of proving that its valid.

• A (quasi-)commonsensical way of proving thisFor the sake of argument, lets call the soccer-playing musician in our class Miles. Now since Miles is a musician, it follows that someone is a musician. So, weve proven our argument.Existential Elimination (E) codifies the reasoning implicit in this passage.

• Existential Elimination, Step 1 of 3Begin with a given statement in which is the main operator.Our example:Somebody in this class is a musician and a soccer player.x (Cx&(Mx&Sx))A

• Step 2 of 3Hypothesize for E by taking the statement from Step 1, removing the , and replacing all instances of the variable associated with with a name that has not been used elsewhere in the derivation.x (Cx&(Mx&Sx))A | Cm & (Mm & Sm)H forE

• Step 3 of 3Derive your desired conclusion, and exit the world of hypothesis by repeating the last line in the world of hypothesis and citing the lines constituting your hypothetical derivation, plus the line in Step 1.

• Example of Step 31. x(Cx&(Mx&Sx))A2. | Cm & (Mm & Sm)H for E3.| Mm & Sm2 &E4.| Mm3 &E5.| xMx4 I6. xMx1, 2-5 EWTP:xMxNotice that all Es stutterbut are otherwise structured like other hypothetical derivations (~I, I)However, they have an extra number in the last line.INSPIRATION ELIMINATION

• Example of Step 31. x(Cx&(Mx&Sx))A2. | Cm & (Mm & Sm)H for E3.| Mm & Sm2 &E4.| Mm3 &E5.| xMx4 I6. xMx1, 2-5 ESomeone in this class is a soccer-playing musicianFor the sake of argument, lets call him MilesSince Miles is a musician.someone is a musician.WTP:xMx

• Special qualifications to the Las Vegas RuleA name used in a hypothesis for E cannot leave the world of hypothesis.So the following is not legitimate:1. x(Cx&(Mx&Sx))A2. | Cm & (Mm & Sm)H for E3.| Mm & Sm2 &E | Mm3 &EMm1,2-4 EThink about what this inference says: Someone in the class is a soccer-playing musician. So Miles is a musician. Clearly invalid!

• Further qualificationsIts also illegitimate to use a name that appears outside of the world of hypothesis in forming your initial hypothesis for E.x(Cx & (Mx & Sx))AMaA |Ca & (Ma & Sa)H for E

• Important word of cautionExistential elimination is probably the trickiest rule to implement in a proof strategy.It doesnt provide easy fodder for reverse engineering.

• An important trickEFQ is really helpful, particularly when you have an E nested inside of a ~I. The way to think about this:In the ~I world of hypothesis, you want a contradictionSo the entire purpose of hypothesizing for E is to get this contradiction.

• Example: Nolt 8.2.7 ~x(Fx&~Fx)1. | x(Fx&~Fx)H for ~I2.||Fa & ~FaH for E3.||Fa2 &E4.||~Fa2 &E5.|| P&~P3,4 EFQ6.| P & ~P1,2-5 E7. ~x(Fx&~Fx)1-6 ~I

• More examples: Nolt 8.2.1x(Fx&Gx) xFx & xGx1. x(Fx&Gx)A2.| Fa & GaH for E3.| Fa2 &E4.| xFx3 I5.| Ga2&E6.| xGx5 I7. | xFx & xGx4,6 &I8. xFx & xGx1, 2-7 E

• Nolt 8.2.3xFx Ga FbxGx1. xFx GaA2.|FbH for I3.|xFx2 I4.|Ga1,3 E5.|xGx4 I6. FbxGx2-5 I

• Nolt 8.2.4x~~Fx xFx1. x~~FxA2.|~~FaH for E3.|Fa2 ~E4.|xFx3 I5.xFx1, 2-4 ENote that you HAVE to use ~E or else you wont have the right kind of lines to trigger E

• Nolt 8.2.8 xFx yFy1. | xFxH for I2.||FaH for E3.||yFy2 I4.|yFy1,2-3 E5. xFx yFy1-4 I6.|yFyH for I7.||FaH for E8.||xFx7 I9.|xFx6,7-8 E10.yFy xFx6-9 I11. xFx yFy5,10 I

• Nolt 8.2.9x(Fx v Gx) xFx v xGx1. x(Fx v Gx)A2. |Fa v GaH for E3.||FaH for I4.||xFx3 I5.||xFx v xGx4 vI6.|Fa (xFx v xGx)3-5 I7.||GaH for I8.||xGx7 I9.||xFx v xGx 8 vI10.|Ga(xFx v xGx)7-9 I11.|xFx v xGx2,6,10 vE12. xFx v xGx1, 2-11 E