ThomasHobbes wrote: ↑October 12th, 2018, 12:30 pm

Mosesquine wrote: ↑October 12th, 2018, 3:27 am

The example that I gave previously is read as:

It is not the case that there exists some x such that x is consciousness, and it is not the case that there exists some y such that y is consciousness, and it is not the case that x is identical to y; and x is an epiphenomenon. The negation ("it is not the case that") is wider than the description ("consciousness"), and the description is narrower than the negation. It's Russellian second occurrence of definite description. I didn't mean 'something else', but I meant 'nothing else'. They are different.

LOL

I'm sure you find that convincing. But you are doing nothing more than repeating yourself. There is no argument here; just logical repetition of an empty unsupported assertion.

There is an argument as follows:

If there exist some x such that x is consciousness, and it is not the case that there exists some y such that y is consciousness, and it is not the case that x is identical to y; and it is not the case that x is an epiphenomenon, then it is not the case that there exist some x such that x is consciousness, and it is not the case that there exists some y such that y is consciousness, and it is not the case that x is identical to y; and x is an epiphenomenon.

There exists some x such that x is consciousness, and it is not the case that there exists some y such that y is consciousness, and it is not the case that x is identical to y; and it is not the case that x is an epiphenomenon.

Therefore, it is not the case that there exists some x such that x is consciousness, and it is not the case that there exists some y such that y is consciousness, and it is not the case that x is identical to y; and x is an epiphenomenon.

It's a simple

modus ponens (if p, then q, and p, then q). The proof of the argument above goes as follows:

1. (∃x)(Fx & ~(∃y)(Fy & x ≠ y) & ~Gx) → ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)

2. (∃x)(Fx & ~(∃y)(Fy & x ≠ y) & ~Gx)

∴ ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)

3. asm: (∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)

4. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx) 1, 2, MP

∴ 5. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx) from 3; 3 contradicts 4.

Q.E.D.

It's a valid argument.