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I just can't wrap my head around why this statement does not hold:

∃x∃yRxy ⊨ ∃x∃yRyx

Let D = {1, 2} and Let I(R) = {(1,2)}

M ⊨ ∃x∃yRxy = 1, since some x(1) has some y(2) such that (x,y) ∈ I(R)

But in my eyes:

M ⊨ ∃x∃yRyx = 1, since some x(2) has some y(1) such that (y,x) ∈ I(R)

Or does this not hold because I(R) = x,y so:

M ⊨ ∃x∃yRyx = 0, since some x(1) has some y(2) such that (y,x) ∈ I(R)

Char
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    It holds: we can always "swap" two adiacent equal quantifiers: $\forall x \forall y$ and $\exists x \exists y$. Thus the second formula is equiv to : $\exists y \exists x Ryx$ and this is the same as the first one. – Mauro ALLEGRANZA Apr 28 '17 at 08:58
  • If $∃x \ ∃y \ Rxy$ holds, this means that there is an "object" $a$ such that $∃y \ Ray$ and this in turn is: there is a $b$ such that $Rab$ holds. But thus there is an "object" $b$ such that $∃y \ Ryb$ holds... – Mauro ALLEGRANZA Apr 28 '17 at 09:01
  • So the book I read had the wrong answer... And here I was pondering why it didn't hold. – Char Apr 28 '17 at 09:03

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