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I'm a beginner in logic and I'm studying with textbooks. Right now I've just got to predicate logic with identity and I need to ask a few questions, so I can free my mind of doubts and sleep well at night.

Do the identity rules (Id)

p//x=x (reflexivity);

x=y ⇔ y=x (symmetry);

x=y/y=z//x=z (transitivity);

Fx/x=y//Fy, Fx/¬Fy//¬(x=y) (substitution);

apply to both variables and constants?

I'm almost certain that they do, but there's this textbook that says they only apply to constants, and then a more recent edition of the same book says it apply to both variables and constants. So I just need to be sure.

Another question:

If I have

  1. Raa
  2. ¬Rab

can I infer from both premises the line ¬(a=b) with the Id rules, or do I need some intermediate step? Or is it just wrong?

One last question: when doing Existential Instantiation (EI), I know I can replace the variable with a new constant, one that did not appear in the proof in any preceding line and in the conclusion line, and then drop the quantifier; but there's a textbook that says I could instantiate with a variable, providing it's a new one that has not been used, so this mean I can do EI with both variables and constants? I was sure that I could only instantiate with a constant, and that the constant was supposed to be a "temporary name". Can anyone clear this to me?

user385442
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  • If you have the rule : "from $Fx$ and $\lnot Fy$ derive : $\lnot (x=y)$", you can apply it with $Rax$ as $Fx$. Thus $Raa$ is $Fa$ and $\lnot Rab$ is $\lnot Fb$ and you can conclude with : $\lnot (a=b)$. – Mauro ALLEGRANZA Nov 03 '16 at 13:53
  • But there is no need to have this rule, because it is a simple consequence of the preceeding one (by tautological equivalence between : $(p \land q) \to r$ and $(p \land \lnot r) \to \lnot q$). – Mauro ALLEGRANZA Nov 03 '16 at 15:32
  • I see, does this mean relational predicates are no different than monadic predicates when we apply the identity rules? Even if it is an intransitive relation? And about the equivalences you mention how do they relate to the identity rules? – user385442 Nov 05 '16 at 06:44

2 Answers2

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I'll treat your first and last question at the same time: there are different formal systems of logic, and yes, some will use variables when eliminating quantifiers, where others use constants. For the former kinds of systems, the identity rules apply to those variables as well as constants, but for the latter types of systems, the identity rules only apply to constants.

Then, to infer $\neg a = b$ you probably need to do a proof by contradiction: assume $a=b$, infer $Rab$ by substituting $b$ for the second $a$ in $Raa$, and that contradicts with $\neg Rab$

Bram28
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  • I see, but I need to know if those rules apply to either variables or constants in systems of natural deduction, like the one I'm using. I just have contradictory information in the textbooks I'm using, and they seem to be about the same system. Also, are you sure there's no way to get to ¬(a=b) with a direct derivation using the Id rules? – user385442 Nov 03 '16 at 00:34
  • Your textbooks are not contradictory; they just define the rules differently. And so what you can or cannot do all depends on how the rules are defined for the particular system you decide to use. So yes, maybe there is a system out there that can get the $\neg a = b $ in one step ... but none that I am familiar with. – Bram28 Nov 03 '16 at 00:40
  • Well, I've checked three different editions of the same textbook, and up to the 11th edition the author was defining the rules of identity saying that they only apply to constants, then on the 12th edition it was saying that they apply to both constants and variables. Perhaps it was a mistake that took him eleven editions to notice. – user385442 Nov 03 '16 at 00:55
  • Perhaps ... though maybe the author changed the system. What in general is highly unlikely though, is that the system would be incomplete, meaning that valid arguments can't be proven to be valid in the system. That would be quite embarrassing to the author, so logicians will take care that their system is complete before they put it in print. So do you think the system up to the 11th edition was incomplete? – Bram28 Nov 03 '16 at 01:03
  • I think it was a mistake. The 11th edition has an exercise where you need to derive x = c, from x = a and a = c, thus the rule is applied to both variables and constants, but he still clearly states in the chapter that the rule applies only to constants. – user385442 Nov 05 '16 at 06:50
  • @Iconoclasteretic Yeah, that looks weird: even if variables are used where other systems would use a constant, it is done within the context of quantifier eliminations, so if this was the problem as a whole, that indeed doesn't look right. What book and author is this? – Bram28 Nov 05 '16 at 17:15
  • It is Hurley's A Concise Introduction to Logic. The difference in the rules to eliminate quantifiers I saw in other books, what I think is a mistake is Hurley's Identity rules up to the 11th edition, he fixed it in the 12th edition though, which is a tad too late. – user385442 Nov 07 '16 at 00:13
  • @Iconoclasteretic I only have Hurley's 10th edition ... Ah, I see! Problem II.10 from section 8.7. Yes, you're absolutely right, in the answer in the back he indeed infers x=c from a=c and x=a even though he says the = rules can only be applied to constants.... which in fact would make his system incomplete: there is no way he can prove that problem without allowing variables to be used for the identity rules. Wow! Good catch!! – Bram28 Nov 07 '16 at 18:00
  • @Iconoclasteretic FYI: The exercise problem did not appear yet in the 7th edition. What did he do in the 12th edition? – Bram28 Nov 07 '16 at 18:23
  • He changed the rule so it can be applied to both variables and constants. Still, can we trust Hurley's? His system is basically Copi's system, slightly dumbed-down. And it still bugs me that different authors employ many different systems of deduction. I can't make my mind on which one is the best to use. The differences in quantifiers' introduction-elimination rules is what annoys me the most. – user385442 Nov 15 '16 at 01:46
  • @Iconoclasteretic Yes, there are lots of different systems ... sometimes you wonder if it is used as a reason to produce yet another book on logic. Personally, I am really not a fan of systems (like Copi and Hurley) where you keep variables while dropping quantifiers; I look at those expressions as formulas with free variables, and hence as not sentences, so I prefer systems where you can only use constants to instantiate quantifiers, and use subproofs to demarcate the role of those constants. – Bram28 Nov 15 '16 at 10:21
  • Yeah, a lot of those more decent logic books, such as the one from Gamut, do it this way. It certainly makes more sense when you think about the formation rules, in the way they're usually defined. My problem is that sometimes it's hard to keep track of those constants you introduce by universal instantiation, if the proof is long, and there are lots of constants, you might generalize on the wrong one. In these systems, you also have a hypothetical rule for existential instantiation, which makes long proofs kind of messy, esp when writing by hand, too many hypotheses to discharge. – user385442 Nov 21 '16 at 01:12
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This answer is to only part of the question. It is similar to what Bram28 suggested although it places the steps in a proof checker:

If I have

  1. Raa
  2. ¬Rab

can I infer from both premises the line ¬(a=b) with the Id rules, or do I need some intermediate step? Or is it just wrong?

Here is a proof:

enter image description here

I can use the assumed equality on line 3 to make the substitution in line 2 which allows me to derive line 4. That leads to a contradiction (line 5) allowing me to negate the assumed equality.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

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Frank Hubeny
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