Do these 4 formulae stand equivalent when symbolizing the statement: 'There is exactly one dog'?
1) ∃x∀y(Dy ↔ y=x)
2) ∃x(Dx ∧ ∀y(Dy → x=y))
3) ∃x(Dx ∧ ¬∃y(¬y=x ∧ Dy))
4) ∃xDx ∧ ∀x∀y((Dx ∧ Dy) → x=y)
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Sign up to join this communityDo these 4 formulae stand equivalent when symbolizing the statement: 'There is exactly one dog'?
1) ∃x∀y(Dy ↔ y=x)
2) ∃x(Dx ∧ ∀y(Dy → x=y))
3) ∃x(Dx ∧ ¬∃y(¬y=x ∧ Dy))
4) ∃xDx ∧ ∀x∀y((Dx ∧ Dy) → x=y)
1) ∃x∀y(Dy ↔ y=x)
There is somethingx such that: everythingy is a dog just in case ity is identical to itx.
2) ∃x(Dx ∧ ∀y(Dy → x=y))
There is a dog and every dog is identical to it.
3) ∃x(Dx ∧ ¬∃y(¬y=x ∧ Dy))
There is a dog and there is nothing else that is a dog.
4) ∃xDx ∧ ∀x∀y((Dx ∧ Dy) → x=y)
There is a dog, and any two things that are dogs are identical.
Here is a sketch of how to prove the equivalence of these four formulae.
(i) ⇒ (ii)
(ii) ⇒ (iv)
(iv) ⇒ (i)
(ii) ⇔ (iii)
— essentially by de Morgan's Law, involving some existential/universal quantifier juggling.