I am trying to convert the unique existential quantifier into plain logic. I have two statements with different placements of the 'for all' quantifier.

∃x ∀y, $(P(y) ∧ P(x)) \rightarrow (y=x)$


∃x, $(P(x) ∧ (∀y, P(y) \rightarrow y=x))$

I'm having trouble figuring out what the difference between these two statements is. The second one I found online and the first one I wrote myself.

If you could let me know the difference, I would greatly appreciate it.

  • $\begingroup$ I don't think you have enough parentheses. In your first statement, don't you mean $\exists x\forall y[(P(y) ∧ P(x)) \rightarrow (y=x)]?$ $\endgroup$ – bof Oct 1 '16 at 7:20
  • $\begingroup$ Suppose there is no $x$ for which $P(x)$ holds. Can you see that in that case the first statement is true? $\endgroup$ – bof Oct 1 '16 at 7:23
  • $\begingroup$ The second sentence is correct if you supply the needed parentheses, that is, change $\forall y P(y)\to y=x$ to $\forall y(P(y)\to y=x).$ Another way to express the unique existential quantifier is $\exists x\forall y(P(y)\leftrightarrow y=x).$ $\endgroup$ – bof Oct 1 '16 at 7:28

The important difference between the two formulas is not in where the universal quantifier is, but that the $P(x)$ is on the left-hand side of the $\to$ in the first version, but outside the entire implication in the second.

Stripping away lesser differences, the structures are $$ (P(x)\land \cdots) \to \cdots \qquad\text{versus}\qquad P(x)\land(\cdots \to \cdots) $$

This means that for the first formula to be true, it demands something only about situations in which $P(x)$ is true -- otherwise the implication is vacuously satisfied. The first formula is automatically true if you find an $x$ for which $P(x)$ doesn't hold.

The second formula wants $P(x)$ to be true in any case and also wants some implication involving $y$s to be true.


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