쿠쿠리 [1310649] · MS 2024 · 쪽지

2025-07-15 13:42:29
조회수 132

이거 문제있음?

게시글 주소: https://orbi.kr/00073840182

R(x) : exists in reality

E(x) : exists

1. ∀x (¬R(x) → E(x))

The negation of 1 is

2. ∃x (¬R(x) ∧ ¬E(x))

2 is a contradiction ("∃x" vs "¬E(x)")

Therefore, 1 is true

Conclusion
∀x (¬R(x) → E(x))

---------------------------------------------------------------------
M(x) : exists in the mind

1. ∀x (¬R(x) → M(x))

The contrapositive of 1 is

2. ∀x (¬M(x) → R(x))

2 is false, therefore 1 is false

The negation of 1 is true

The negation of 1 is

3. ∃x (¬R(x) ∧ ¬M(x))

Conclusion

∃x (¬R(x) ∧ ¬M(x))

------------------------------------------------------------------

1. ∀x (E(x))

The negation of 1 is

2. ∃x (¬E(x))

2 is a contradiction ("∃x" vs "¬E(x)")

Therefore, 1 is true

Conclusion

∀x (E(x))

∀x (E(x)) and ¬∃x (¬E(x)) are equivalent.

The meaning of ¬∃x (¬E(x)) is "Something that does not exist does not exist".

Since the negation of ∀x (E(x)) is a contradiction,

then ∀x (E(x)) is a tautology,

and therefore ∀x (E(x)) is true regardless of the domain of x.

Absolutely anything exists




예전에 썼던건데 지금보니까 신기함

0 XDK (+0)

  1. 유익한 글을 읽었다면 작성자에게 XDK를 선물하세요.