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Context : find a counter example to the statement $$∀x∈C(∃y∈C(y∈A)⇒¬(x∈B))$$

Suppose I know that there is some x such that : x belongs to set C but NOT to set B.

Am I allowed to derive from this that :

for all y (y does not belong to C OR y does not belong to A OR x does not belong to B)

It seems to me that " x does not belong to B" being true, that makes the whole disjunction true for any variable ( even if I have no information on this variable coming from elsewhere).

Is there an official rule that makes this derivation legal in predicate logic natural deduction?

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    If $x$ is "whatever" (i.e. nothing has been assumed about it, yes the move is legal; it is called Universal Generalization. – Mauro ALLEGRANZA Oct 31 '19 at 11:53
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    But to assume that "there is some x such that : x belongs to set C but NOT to set B" dose not implies that this will be true for every $x$ whatever. The UG rule is correctly applied if we have proved, for an $x$ whatever, that "x belongs to set C but NOT to set B". – Mauro ALLEGRANZA Oct 31 '19 at 11:55
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    Write it in formulas: it is simpler. You want to derive $\forall y [\lnot C(y) \lor \lnot A(y) \lor \lnot B(x)]$. Your assumption is $(C(x) \land \lnot B(x))$, for some $x$. The trick is "for some"; if you mean $\exists x$, the derivation is not correct. – Mauro ALLEGRANZA Oct 31 '19 at 12:07
  • @Mauro Allegranza. In the sentence I want to derive, the universal quantification bares on a new variable y . Does this make a difference? –  Oct 31 '19 at 12:09
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    If instead you have proved $C(x) \land \lnot B(x)$, for $x$ whatever, this means $\forall x (C(x) \land \lnot B(x))$, form which $\forall x \lnot B(x)$. Now, from $B(x)$ we have $B(x) \lor \varphi (y)$ for $\varphi$ whatever, and then we can generalize on $y$. – Mauro ALLEGRANZA Oct 31 '19 at 12:09

1 Answers1

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Context : find a counter example to the statement $$∀x∈C(∃y∈C(y∈A)⇒¬(x∈B))$$

It seems to me that " x does not belong to B" being true, that makes the whole disjunction true for any variable ( even if I have no information on this variable coming from elsewhere).

Indeed, when $C$ and $B$ are disjoint sets, then the statement would be true.   So that can not serve as a counterexample.

Your approach is no good.   You should seek an interpretation of the sets which makes the statement false.

Now, recall that an implication is false only when the antecedent is true but the consequent is false.

So you want $A,B,C$ to be such that the negation of the statement, ie $\exists x\in C~((\exists y\in C~(y\in A))\wedge(x\in B))$, is true.

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Graham Kemp
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