Proving dogs exist
If x barks then x is a dog: ∀x(Bx → Dx)
t: Timmy (a dog)
PROOF:
- ∀x(Bx → Dx) [Premise]
- Bt [Premise]
- Bt → Dt [1 UI]
- Dt [2, 3 MP]
- ∃x(Dx) [4 EG]
QED
Proving ghosts don't exist
If the Hessian rider is imperceptible, then ghosts don't exist: ¬Ph → ¬∃x(Gx)
Ph: The headless Hessian rider (Sleepy Hollow) is perceptible
PROOF:
- ¬Ph → ¬∃x(Gx)[Premise]
- ¬Ph [Premise]
- ¬∃x(Gx) [1, 2 MP]
¬∃x(Gx) ⇔ ∀x(¬Gx)
QED
Here I faced considerable difficulty. Observe the categorical logic syllogism below:
- All imperceptible thingsare nonexistent things
- All ghosts are imperceptible things
Ergo, - All ghosts are nonexistent things
QED
No problem at all, but issues pop up when translating the above syllogism into predicate logic. It's difficult to render premise 1 in predicate logic. How should I end this sentence, ∀x(¬Px → ???)?. Existence is not a predicate in predicate logic and if so, I have nothing to negate, to put where ??? is.
If the conclusion is No A are B (no ghosts are existent things) then, we would have to use a particular affirmative statement, viz. ∃x(Gx), there is at least one ghost (that exists) to negate.
The problem with this "proof" is that the Hessian rider is just one instance of a ghost that's imperceptible. To generalize the imperceptibility and thus nonexistence to all ghosts is fallacious. We need an all statement, I can't for the life of me come up with one. Help! Comments/suggestion/answers welcome.
We could do the following: Imagine there are exactly 2 ghosts claimed to exist, Myrtle (m) and the Hessian (h). Then the following would be a valid/sound argument.
PROOF:
- ¬Pm ∧ ¬Ph
- (¬Pm ∧ ¬Ph) → ¬∃x(Gx)
- ¬∃x(Gx)
QED
Suppressed premise: ∀x∀y∀z((Gx ∧ Gy ∧ Gz) → x = y ∧ x ≠ z). There are exactly 2 ghosts.
Derived from the above technique, these statements ∀x(Gx → ¬Px) → ¬∃x(Gx): If for all x, x a ghost THEN x is imperceptible IMPLIES Ghosts don't exist AND ∀x((Gx → ¬Px) → ¬∃x(Gx)), looked full of promise, but it leads to a contradiction because if you instantiate Gx, which you must, you end up with ∃x(Gx) ∧ ¬∃x(Gx), a contradiction.
Proving earth exists
Pe = Earth is perceptible
There exists a thing and that thing is identical to earth: ∃x(x = e) i.e. earth exists
PROOF:
- Pe → ∃x(x = e)[Premise]
- Pe [Premise]
- ∃x(x = e) [1, 2 MP]
QED
Not sure if ∃x(x = e) is the correct translation for earth exists. We're talking about 1 individual here and so it's easier to prove.
Proving Santa does not exist
Exists a thing that is Santa: ∃x(x = s)
Santa is perceptible: Ps
PROOF:
- ¬Ps → ¬∃x(x = s) [Premise]
- ¬Ps [Premise]
- ¬∃x(x = s) [1, 2 MP]
QED
1 individual person and so again easier (to prove existence). Here too, unsure whether ¬∃x(x = s) is the correct translation for Santa Claus does not exist. It feels right, it means nothing is Santa.
SUMMARY:
The following were easier to prove:
- Some dogs exist (Particular affirmative)
- Earth exists (An individual, a constant)
- Santa does not exist (An individual, a constant)
The following remain unresolved (for me):
- No ghosts exist (Universal negative)
Also, I have the same exact issue with UNIVERSAL AFFIRMATIVE claims about existence. How do I prove/say All humans exist? ∀x(Hx → ???), where Hx: x is human. Do I do the same thing I did with the universal negative?