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 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, 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 & Sm2 &E 4.| Mm3 &E 5.|  xMx4  I 6.  xMx1, 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. INSPIRATION  ELIMINATION

Example of Step 3 1.  x(Cx&(Mx&Sx))A 2. | Cm & (Mm & Sm)H for  E 3.| Mm & Sm2 &E 4.| Mm3 &E 5.|  xMx4  I 6.  xMx1, 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))A 2. | Cm & (Mm & Sm)H for  E 3.| Mm & Sm2 &E 4. | Mm3 &E 5.Mm1,2-4  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.MaA 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. 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 ├ ~  x(Fx&~Fx) 1. |  x(Fx&~Fx)H for ~I 2.||Fa & ~FaH for  E 3.||Fa2 &E 4.||~Fa2 &E 5.|| P&~P3,4 EFQ 6.| P & ~P1,2-5  E 7. ~  x(Fx&~Fx)1-6 ~I

More examples: Nolt  x(Fx&Gx) ├  xFx &  xGx 1.  x(Fx&Gx)A 2.| Fa & GaH for  E 3.| Fa2 &E 4.|  xFx3  I 5.| Ga2&E 6.|  xGx5  I 7. |  xFx &  xGx4,6 &I 8.  xFx &  xGx1, 2-7  E

Nolt  xFx → Ga ├ Fb→  xGx 1.  xFx → GaA 2.|FbH for →I 3.|  xFx2  I 4.|Ga1,3 →E 5.|  xGx4  I 6. Fb→  xGx2-5 →I

Nolt  x~~Fx ├  xFx 1.  x~~FxA 2.|~~FaH for  E 3.|Fa2 ~E 4.|  xFx3  I 5.  xFx1, 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 ├  xFx ↔  yFy 1. |  xFxH for →I 2.||FaH for  E 3.||  yFy2  I 4.|  yFy1,2-3  E 5.  xFx →  yFy1-4 →I 6.|  yFyH for →I 7.||FaH for  E 8.||  xFx7  I 9.|  xFx6,7-8  E 10.  yFy →  xFx6-9 →I 11.  xFx ↔  yFy5,10  I

Nolt  x(Fx v Gx) ├  xFx v  xGx 1.  x(Fx v Gx)A 2. |Fa v GaH for  E 3.||FaH for →I 4.||  xFx3  I 5.||  xFx v  xGx4 vI 6.|Fa →(  xFx v  xGx)3-5 →I 7.||GaH for →I 8.||  xGx7  I 9.||  xFx v  xGx 8 vI 10.|Ga→(  xFx v  xGx)7-9 →I 11.|  xFx v  xGx2,6,10 vE 12.  xFx v  xGx1, 2-11  E