- #1
Aziza
- 190
- 1
According to my book,
[itex](\exists!x)P(x)[/itex] is equivalent to [itex](\exists x)P(x)\wedge(\forall y)(\forall z)[P(y)\wedge P(z)\Rightarrow y=z][/itex]
But I don't see why the variable z is necessary. Wouldn't the following also be correct but shorter and easier to understand:
[itex](\exists x)P(x)\wedge(\forall y)(P(y)\Rightarrow y=x)[/itex]
??
[itex](\exists!x)P(x)[/itex] is equivalent to [itex](\exists x)P(x)\wedge(\forall y)(\forall z)[P(y)\wedge P(z)\Rightarrow y=z][/itex]
But I don't see why the variable z is necessary. Wouldn't the following also be correct but shorter and easier to understand:
[itex](\exists x)P(x)\wedge(\forall y)(P(y)\Rightarrow y=x)[/itex]
??