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Let M(x, y) be the statement “x is a mother of y”, where the universe of discourse is the set of all people in the world.

Express the statement "A person’s mother’s mother cannot be his/her mother."

I have come up with two solutions and they seem to be the same to be. I can't see the differences in results for both.

  1. ∀x∃y∃z(M(y,x)∧ M(z,y) ∧ ¬M(z,x)
  2. ∀x∃y∃z((M(y,x)∧M(z,y)) -> ¬M(z,x))
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  • $\begingroup$ The negation of your statement is that a person's mother's mother can be his/her mother, which is $M(y,x)\wedge M(z,y)\wedge M(z,x)$. This is not the negation of the formula inside of the quantifiers in (1). $\endgroup$ – Christopher Carl Heckman Sep 29 '15 at 6:38
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Both are incorrect for different reasons.

The correct answer is: $$\forall x\;\forall y\;\forall z \;\Big(\big( M(y,x) \land M(z,y) \big)\Rightarrow\neg M(z,x)\Big)$$

For ease of explanation, I will work with the following slightly changed statement (because it sounds nicer)

"My mother's mother cannot be my mother"

This can be rewritten

"If you are my mother's mother, then you are not my mother"

(1) is incorrect. The reason for this is that (1) being true as written does not preclude that for a given x there exist j,k such that

$$M(j,x) \land M(k,j) \land M(k,x)$$

(2) as written does not preclude the above statement either either. Think about it, all you need to do is find a $k$ such that $M(k,x)$ is false, and then your implication is trivially true.

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