It is well-known that $(x+1)(x-1)=0 \iff x=-1~\text{or}~x=1$. Does this mean that the open sentences $(x+1)(x-1)=0$ and $x=-1~\text{or}~x=1$ are equivalent because they have the same truth set $\{-1,1\}$?
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Your reasoning makes sense (assuming the universe is a domain. Do you doubt something? – Karl Dec 22 '21 at 05:23
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Thank you for asking. Not really. I'm just trying to figure out a reason behind the equivalence. – KHOOS Dec 22 '21 at 06:39
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1This question is best to be transferred to Philosophy SE; it is related to a question what Kit Fine calls "antinomy of the variable" in his The Role of Variables. – Tankut Beygu Dec 26 '21 at 08:27
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Given two formulas $\phi,\psi$ with free variables in $X$ with $x\in X$ taking a value in the set $D_x$, they are equivalent if and only if for any valuation $\sigma : X \to \bigcup_{x\in X}D_x$ (that is $\sigma x \in D_x$), $\phi_\sigma \iff \psi_\sigma$, where $\xi_\sigma$ is $\xi$ in which each variable $x$ has been substituted by its value $\sigma x$.
Couchy
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Is this answer actually saying that they are equivalent because $\forall x \in \mathbb{R}: [(x+1)(x-1) = 0 \Leftrightarrow (x = -1 \lor x = 1)]$? – Hermis14 Dec 22 '21 at 11:56
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The equivalence of $(x+1)(x-1)=0\tag*{}$ and $x=-1\:\:\text{or}\:\: x=1\tag*{}$(typically in the context of solving equations) is tacitly universally quantified $\forall x\;\Big((x+1)(x-1)=0\iff \big(x=-1\:\:\text{or}\:\:x=1\big)\Big),\tag*{}$ and indeed means that they have the same truth set.
On the other hand, the equivalence of $\forall x\;\;(x+1)(x-1)=0\tag*{}$ and $\forall x\;\big(x=-1\:\:\text{or}\:\: x=1\big)\tag*{}$ means that they have the same truth value. This is logically weaker than the previous assertion.
ryang
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May I ask, what is the reason behind the writing $(x-1)(x+1)=0 \iff x=1~\text{or}~x=-1$, since it appears rather commonly. Does it just mean that the truth sets are the same? – KHOOS Dec 24 '21 at 00:46
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