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