Midterm coming up later today. Don't understand how in a) the formula doesn't change. I know that x+y is in the scope of Q(x,y,z) but then we're suppose to introduce a new variable and substitute, right?
Please explain please!!
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See answer to the similar post: substitution-for-bound-variables-in-logic – Mauro ALLEGRANZA Jun 08 '17 at 13:13
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Regarding (c), I agree with you; you have to check the fine details of the formal specifications on your textbook, in order to verify why the "renaming" of $y$ is allowed and that of $x$ is not. – Mauro ALLEGRANZA Jun 09 '17 at 08:22
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Hint: The notation $\varphi[t/x]$ denotes that in the formula $\varphi$ all free occurrences of the variable $x$ are replaced by $t$, possibly involving renaming to avoid name clashes. Is $x$ free in the $\varphi$ that is mentioned?
Hans Hüttel
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Ah I get it now. For a), all x's are bounded, so there's no free occurrence of x, thus we don't substitute. – JiHua Jun 08 '17 at 13:19
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For c) though, I understand that a new variable u is introduced to deal with the y in f(x+y,x), but why didn't we introduce another variable for x? – JiHua Jun 08 '17 at 13:20
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See Herbert Enderton, A Mathematical Introduction to Logic, Academic Press (2nd ed., 2001), page 112, for details.
We need a formal definition of substitution:
For atomic $α$, $α[t/x]$ is the expression obtained from $α$ by replacing the variable $x$ by $t$.
Obvious clauses for the connectives.
$(∀y \ α)[t/x] = ∀y \ α$, if $x = y, \ \ \ $ and $ \ \ \ = ∀y \ (α[t/x])$, if $x \ne y$; and the same for $\exists$.
Thus, in case (a) above, $x=y$ in $\varphi := \exists x \ \alpha(x)$ and the result is: $\exists x \ \alpha(x)$.
Obviously, we can apply the technique of "alphabetic variant" [see Enderton, page 126] to get a formula $\varphi'$, which differs from $\varphi$ only in the choice of quantified variables, i.e. $\varphi' := \exists w \ \alpha(w)$, such that $\varphi$ and $\varphi'$ are equivalent and $t$ is substitutable for [see page 113] $x$ in $\varphi'$.
But in this case the substitution will be: $\varphi' [t/x]$.
Mauro ALLEGRANZA
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