Can we derive this from =Elim rule?

Question

I know we can derive something like this from =Elim rule: $$\varphi(a,b)$$ $$b=c$$ $$\therefore\varphi(a,c//b)$$ But I want to know if we can also go ahead and derive something like this if there was some need to do so: $$\varphi(a,c//b)$$ $$a=d$$ $$\therefore\varphi(d//a,c//b)$$ I tried to find a sample of such derivation with no success.

Answer 1

Your question points the finger to a little detail not always clearly stated.

See : Dirk van Dalen, Logic and Structure, Springer (5th ed - 2013), page 93 :

$$\frac { x_1 = y_1, \ldots, x_n = y_n \quad \quad \quad ϕ(x_1,\ldots, x_n) } { ϕ(y_1,\ldots, y_n)}$$

where $y_1,\ldots, y_n$ are free for $x_1,\ldots, x_n$ in $ϕ$.

Note that we want to allow substitution of the variable $y_i (i ≤ n)$ for some and not necessarily all occurrences of the variable $x_i$.

If we want to formulate the rule using the simultaneous substitution operator, we must have :

$$\frac { x_1 = y_1, \ldots, x_n = y_n \quad \quad \quad ϕ[x_1,\ldots, x_n/z_1,\ldots,z_n] } { ϕ[y_1,\ldots, y_n/z_1,\ldots,z_n]}.$$

Compare it with a textbook based on Natural Deduction: Ian Chiswell & Wilfrid Hodges, Mathematical Logic (2007).

The $(= \text E)$ rule is defined [ page 122 ] as follows :

If $\phi$ is a formula, $s$ and $t$ are terms substitutable for $x$ in $\phi$, and $D ... (s = t)$, $D' ... \phi[s/x]$ are derivations, then so is

$$\frac { D ... (s = t) \quad \quad \quad D' ... \phi[s/x] } { \phi[t/x]}$$

without further "comments" on $\phi[t/x]$.

The operation of substitution for FOL expressions is defined [ page 110 ] as follows:

Definition 5.2.10 Suppose $E$ is an expression, $y_1,\ldots , y_n$ are distinct variables and $t_1,\ldots, t_n$ are terms such that each $t_i$ is substitutable for $y_i$ in $E$. Then we write

$E[t_1/y_1,\ldots , t_n/y_n]$

for the expression got from $E$ by simultaneously replacing each [emphasis mine] free occurrence of each $y_i$ in $E$ by $t_i$.

The "trick" (see the derivation of symmetry of $=$, page 122) is the suitable choice if $\phi$.

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We can see also the Hilbert-style version of equality axioms in:

• Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001), page 112:

$x = y → (α → α')$

where $α$ is atomic and $α'$ is obtained from $α$ by replacing $x$ in zero or more (but not necessarily all) [emphasis mine] places by $y$,

where the use of the symbolism for substitution ( $\alpha^x_t$ in Enderton's book) is carefully avoided.

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Having said that, here are some legitimate applications of $(= \text E)$:

$\dfrac {x =y \quad \quad \quad x^2 + y^2 > 12x }{ 2y^2 > 12x}(\text {partial substitution})$

$\dfrac {x =y \quad \quad \quad x^2 + y^2 > 12x }{ x^2 + y^2 > 12x}(\text {no substitution})$

$\dfrac {x =y \quad \quad \quad x^2 + y^2 > 12x }{ 2y^2 > 12y}(\text {total substitution})$

Answer 2

The derivation makes sense, but the notation is a little confusing. To be specific, while it does make sense in the first inference to write $\varphi(a,c//b)$ to indicate that the conclusion is a statement that results from replacing any (which may end up being all, some, or none) of the $b$'s with $c$'s, to use this notation as a premise in the second inference is a little weird, because the premise is something you have, rather than something you end up with. Indeed, if we were to replace different $b$'s with $c$'s in $\varphi(a,c//b)$ as compared to which ones we replace in $\varphi(d//a,c//b)$, then the inference is invalid!

So, given that in the second inference you are in fact leaving any of the $b$'s (and $c$'s) alone, you really shouldn't use this substitution notation for them, but use it just for the substitition of $d$'s for $a$'s, as that is what you are doing. Hence, as Mario said, the second inference is really:

$\varphi(a,c)$

$a=d$

$\therefore \varphi(d//a,c)$