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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.
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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. • x(Cx&(Mx&Sx))├ xMx • This is clearly valid, yet we don’t have a way of proving that it’s valid.
A (quasi-)commonsensical way of proving this… • For the sake of argument, let’s call the soccer-playing musician in our class Miles. Now since Miles is a musician, it follows that someone is a musician. So, we’ve proven our argument. • Existential Elimination (E) codifies the reasoning implicit in this passage.
Begin 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 Existential Elimination, Step 1 of 3
Step 2 of 3 • Hypothesize for E bytakingthestatementfromStep 1, removingthe , andreplacingallinstancesofthevariableassociatedwith with a namethathasnotbeenusedelsewhere in thederivation. • x (Cx&(Mx&Sx)) A • | Cm & (Mm & Sm) H for E
Step 3 of 3 • Derive 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, plusthe line in Step 1.
Example of Step 3 WTP: xMx INSPIRATION 1. x(Cx&(Mx&Sx)) A 2. | Cm & (Mm & Sm) H for E 3. | Mm & Sm 2 &E 4. | Mm 3 &E 5. | xMx 4 I 6. xMx 1, 2-5 E DERIVATION However, they have an extra number in the last line. Notice that all E’s “stutter” …but are otherwise structured like other hypothetical derivations (~I, I) ELIMINATION
Example of Step 3 WTP: xMx Someone in this class is a soccer-playing musician 1. x(Cx&(Mx&Sx)) A 2. | Cm & (Mm & Sm) H for E 3. | Mm & Sm 2 &E 4. | Mm 3 &E 5. | xMx 4 I 6. xMx 1, 2-5 E For the sake of argument, let’s call him Miles Since Miles is a musician… ….someone is a musician.
Special qualifications to the Las Vegas Rule… • A name used in a hypothesis for E cannot leave the world of hypothesis. • So the following is not legitimate: 1. x(Cx&(Mx&Sx)) A 2. | Cm & (Mm & Sm) H for E 3. | Mm & Sm 2 &E • | Mm 3 &E • Mm 1,2-4 E Thinkabout what this inference says: Someone in the class is a soccer-playing musician. So Miles is a musician. Clearly invalid!
Further qualifications… • It’s 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)) A • Ma A • |Ca & (Ma & Sa) H for E
Important word of caution • Existential elimination is probably the trickiest rule to implement in a proof strategy. • It doesn’t provide easy fodder for reverse engineering.
An important trick… • EFQ 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 contradiction • So 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 ~I 2. | |Fa & ~Fa H for E 3. | |Fa 2 &E 4. | |~Fa 2 &E 5. | | P&~P 3,4 EFQ 6. | P & ~P 1,2-5 E 7. ~x(Fx&~Fx) 1-6 ~I
More examples: Nolt 8.2.1 x(Fx&Gx) ├ xFx & xGx 1. x(Fx&Gx) A 2. | Fa & Ga H for E 3. | Fa 2 &E 4. | xFx 3 I 5. | Ga 2&E 6. | xGx 5 I 7. | xFx & xGx 4,6 &I 8. xFx & xGx 1, 2-7 E
Nolt 8.2.3 xFx → Ga ├ Fb→xGx 1. xFx → Ga A 2. |Fb H for →I 3. |xFx 2 I 4. |Ga 1,3 →E 5. |xGx 4 I 6. Fb→xGx 2-5 →I
Nolt 8.2.4 x~~Fx ├ xFx 1. x~~Fx A 2. |~~Fa H for E 3. |Fa 2 ~E 4. |xFx 3 I 5.xFx 1, 2-4 E Note that you HAVE to use ~E or else you won’t have the right kind of lines to trigger E
Nolt 8.2.8 ├ xFx ↔ yFy 1. | xFx H for →I 2. | |Fa H for E 3. | |yFy 2 I 4. |yFy 1,2-3 E 5. xFx → yFy 1-4 →I 6. |yFy H for →I 7. | |Fa H for E 8. | |xFx 7 I 9. |xFx 6,7-8 E 10.yFy → xFx 6-9 →I 11. xFx ↔ yFy 5,10 I
Nolt 8.2.9 x(Fx v Gx) ├ xFx v xGx 1. x(Fx v Gx) A 2. |Fa v Ga H for E 3. | |Fa H for →I 4. | |xFx 3 I 5. | |xFx v xGx 4 vI 6. |Fa →(xFx v xGx) 3-5 →I 7. | |Ga H for →I 8. | |xGx 7 I 9. | |xFx v xGx 8 vI 10. |Ga→(xFx v xGx) 7-9 →I 11. |xFx v xGx 2,6,10 vE 12. xFx v xGx 1, 2-11 E