Existential Elimination Kareem Khalifa Department of Philosophy Middlebury College
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Existential Elimination Kareem Khalifa Department of Philosophy Middlebury College

Jan 12, 2016

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Transcript

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.

Existential Elimination, Step 1 of 3

• Begin with a given statement in which is the main operator.

• Our example:• Somebody in this class is a musician and a

soccer player.

1. x (Cx&(Mx&Sx)) A

Step 2 of 3

• Hypothesize 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.

1. x (Cx&(Mx&Sx)) A

2. | Cm & (Mm & Sm) H forE

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, plus the line in Step 1.

Example of Step 3

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

WTP:xMx

Notice that all E’s “stutter”

…but are otherwise structured like other hypothetical derivations (~I, I)

However, they have an extra number in the last line.

INSPIRATIONINSPIRATION

ELIMINATION ELIMINATION

Example of Step 3

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

Someone in this class is a soccer-playing musician

For the sake of argument, let’s call him Miles

Since Miles is a musician…

….someone is a musician.

WTP:xMx

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)) A2. | Cm & (Mm & Sm) H for

E3. | Mm & Sm 2 &E4. | Mm 3 &E5. Mm 1,2-4 E

Think about 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.

1. x(Cx & (Mx & Sx)) A

2. Ma A

3. |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.

contradiction– So the entire purpose of hypothesizing for E

Example: Nolt 8.2.7

├ ~x(Fx&~Fx)1. | x(Fx&~Fx) H for ~I2. | |Fa & ~Fa H for E3. | |Fa 2 &E4. | |~Fa 2 &E5. | | P&~P 3,4 EFQ6. | P & ~P 1,2-5 E7. ~x(Fx&~Fx) 1-6 ~I

More examples: Nolt 8.2.1

x(Fx&Gx) ├ xFx & xGx1. x(Fx&Gx) A2. | Fa & Ga H for E3. | Fa 2 &E4. | xFx 3 I5. | Ga 2&E6. | xGx 5 I7. | xFx & xGx 4,6 &I8. 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 ENote 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 ↔ yFy1. | xFx H for →I2. | |Fa H for E3. | |yFy 2 I4. |yFy 1,2-3 E5. xFx → yFy 1-4 →I6. |yFy H for →I7. | |Fa H for E8. | |xFx 7 I9. |xFx 6,7-8 E10.yFy → xFx 6-9 →I11. xFx ↔ yFy 5,10 I

Nolt 8.2.9x(Fx v Gx) ├ xFx v xGx1. x(Fx v Gx) A2. |Fa v Ga H for E3. | |Fa H for →I4. | |xFx 3 I5. | |xFx v xGx 4 vI6. |Fa →(xFx v xGx) 3-5 →I7. | |Ga H for →I8. | |xGx 7 I9. | |xFx v xGx 8 vI10. |Ga→(xFx v xGx) 7-9 →I11. |xFx v xGx 2,6,10 vE12. xFx v xGx 1, 2-11 E

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