3

So we have three statements: (i), (ii), and (iii) and we want to prove they are equivalent. In the notes that I am reading, the author proved (i) $\implies$ (iii), (iii) $\implies$ (ii), (iii) $\implies$ (i) and finally (ii) $\implies$ (i). However, during the proof of (ii) $\implies$ (i), the author had to also use the result of (iii) $\implies$ (i). I want to ask is this allowed? I thought we were supposed to assume (ii) to prove (i) without making use of the prior results.

Xander Henderson
  • 29,772
Bill
  • 4,417
  • I don't see a problem, you also don't have to assume $(ii)$ when proving $(ii)\implies (i)$, there are other methods as well (but assuming $(ii)$ and showing that $(i)$ holds is fine of course). And after all, you could just copy the proof from $(iii)\implies(i)$ into the proof for $(ii)\implies(i)$, it would not change the logical content but just be more tedious to read. – fweth Jul 18 '22 at 06:19
  • It is weird that the implication $3-1$ was proved before $2-1$ but since it was proven before, it is OK to use it for further results. To really judge whether the proof is correct, we will need to see the concrete proof. – Peter Jul 18 '22 at 06:19
  • Yes that's allowed. That's why they have done $ (iii) \implies(i)$ first before $(ii) \implies (i)$. The usual way is $(i) \implies (iii)$, $(iii) \implies (ii)$, $(ii) \implies (i)$ but if $(ii) \implies (i)$ can't be reached directly we can use $(i) \implies (iii)$, $(iii) \implies (ii)$, ($(iii) \implies (i)$ ,$(ii) \implies (i)$). – Dan Jul 18 '22 at 06:21
  • Ah, now I see what might be confusing you. If I prove $(iii)\implies(i)$ by saying "Let's assume $(iii)$ and then [PROOF]" then you can't use [PROOF] in the proof of $(ii)\implies(i)$ because you can't assume $(iii)$. But when they refer to the proof of $(ii)\implies(i)$, they would use the complete statement, which makes no assumptions, and hence is applicable everywhere. – fweth Jul 18 '22 at 06:24
  • You are right however that the implications $3-1$ and $3-2$ are not enough for the implication $2-1$. But I think, $2$ is assumed aditionally to $3-1$ and if we then arrive at $1$ , everything is OK. – Peter Jul 18 '22 at 06:28
  • @Dan , hi. Can you elaborate a bit on what you mean by :(i)⟹(iii) , (iii)⟹(ii), ((iii)⟹(i) ,(ii)⟹(i)). Especially the last part ((iii)⟹(i) ,(ii)⟹(i)). – Bill Jul 18 '22 at 06:54
  • [disappearing comment] @Bill I added 2 new links to my Answer below. – ryang Jul 18 '22 at 09:37

3 Answers3

8

Summarising your description: the author has shown that

  1. $A \implies C,$
  2. $C \implies B,$
  3. $C \implies A,$
  4. $B \implies A,$ using result #3.

That is, the author has proven that $$(A→C)∧(C→B)∧(C→A)∧\Big((C→A)→(B→A)\Big),$$ which, via Modus Ponens, logically entails that $$(A→C)∧(C→B)∧(B→A),$$ which logically entails that $$(A↔B)∧(A↔C)∧(B↔C),$$ which means that $A,B,C$ are logically equivalent, as required.

(In fact, the two logical entailments above are actually logical equivalences, but here, this stronger assertion is unnecessary.)

Hence, the author's proof is a perfectly valid.

P.S. For interest, point #4 of my previous answer Proof of multiple equivalences contains alternative suggestions for proving that $A,B,C$ are logically equivalent.

P.P.S. It's worth pointing out that the sentence $$A↔B↔C,$$ strictly logically speaking, does not mean that $A,B,C$ are logically equivalent: it is not true when $A,B,C$ are all false. My explanation here: Associativity of logical connectives

ryang
  • 38,879
  • 14
  • 81
  • 179
4

Why not? Let's call the three statements A, B and C. Now the author is stating and proving a few lemmas:

Lemma 1. If A then C

Lemma 2. If C then B

Lemma 3. If C then A

Lemma 4. If B then A

In the proof of Lemma 4, the author makes use of Lemma 3, which is obviously allowed because Lemma 3 has already been proved. Now the author states

Theorem. The following conditions are equivalent:
(i) A
(ii) B
(iii) C

Proof. (i) implies (iii) is Lemma 1; (iii) implies (ii) is Lemma 2; (ii) implies (i) is Lemma 4. QED

egreg
  • 238,574
1

Then the author's proof must have been redundant. Let's say it has already been proved that $(i) \implies (iii)$, $(iii) \implies (i)$ and $(iii) \implies (ii)$. Now to prove that all statements are equivalent, you need to prove one of the following:

  • $(ii) \implies (iii)$
  • $(ii) \implies (i)$

Now, by your claim the proof of $(ii) \implies (i)$ looks something like: $$(ii) \implies ... \implies (iii) \implies (i)$$ Notice that stopping at $(iii)$ would have been enough.

eipim1
  • 123
  • but we did not prove 2-3. Yes we would be complete if we can prove 2-1, but the proof of 2-1 itself uses the result of 3-1. – Bill Jul 18 '22 at 08:31
  • I am not saying that the proof is wrong. I am just saying that if $(ii) \implies (iii)$ then you do not need to go further another step to say that okay we proved $(ii) \implies (i)$ – eipim1 Jul 18 '22 at 08:50
  • "by your claim, the proof of $B \implies A$ looks something like $B \implies ... \implies C \implies A$" The OP didn't suggest this particular possibility though. – ryang Jul 18 '22 at 10:26