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.