1

The reason for posting this question is a slightly different approach being used for the proof.

If $a\mid b$, & $a\nmid c$, then $b\nmid c$.
I have used the contradiction approach:

$\exists k \in \mathbb{Z}, b=ak$. If assume that $b \mid c$, then $\exists m \in \mathbb {Z}, c=bm$.
This means that $c=akm$.
But if $\forall l \in \mathbb{Z}, c \ne al$, then this is proved wrong, as $l=km$. Hence, there can be no integer value of $l=km$ possible.

jitender
  • 707
  • 1
    I think you ended before the last lime. Once you've got $c=akm$, $a|c$, so it's not true that $a\not| c$, and hence $\neg (a|b&a\not|c)$. – Javi Feb 15 '18 at 21:48
  • @Javi It seems that you have negated the hypothesis, with the 'AND' between the two simple propositions in the compound proposition in the hypothesis. – jitender Feb 15 '18 at 21:54
  • 1
    $(P\land Q)\Rightarrow R\equiv \neg R\Rightarrow \neg (P\land Q)$. I thought you were assuming $\neg R$ ($b|c$) and then, assuming $(P\land Q)$ to come to a contradiction, which would lead to $\neg (P\land Q)$. – Javi Feb 15 '18 at 21:57
  • 1
    Yes Jami, and $\lnot (P\land Q) \equiv \lnot P \lor \lnot Q$ So the asker has correctly shown that assuming the premises and the negation of the conclusion leads to the negation of $a \nmid c$, and hence $\lnot (a\mid b \land a\nmid c)$ – amWhy Feb 15 '18 at 22:07
  • @amWhy You seem to point to the contra-positive approach, with negating hypothesis and conclusion, with places exchanged (i.e., conclusion being hypothesis, & vice-versa). If I am correct, then the contradiction approach is taken by using the contra-positive. – jitender Feb 15 '18 at 22:17
  • 1
    Um.... what's your question? The proof is valid and complete. – fleablood Feb 15 '18 at 22:21
  • 1
    Contrapositive approach is fine. We have proven till we are sick in the face that if $a|b$ and $b|c$ then $a|c$. Or in other words: given $a|b$ then $b|c\implies a|c$. The contrapositive (which is always equivalent) is: give $a|b$ then $\lnot a|c \implies \lnot b|c$. So that's that, we are done. The contrapositives are equivalent statement so those are equivalent (true) statements. – fleablood Feb 15 '18 at 22:26
  • 1
    jNo, in your case, we need both the premise (which I have not negated). The negation of the premise comes by assuming the truth of the premise, seeing then what the negation of the conclusion leads us, and in that way arrive at a contradiction. In particular, we obtain a contradiction only assuming both the premise, and the negation of the conclusion. You would not have arrived at your contradiction without having implicitly assumed the premise (antecedent). So it is in fact a proof by contradiction. – amWhy Feb 15 '18 at 22:51

1 Answers1

2

Yes, you've successfully proven the implication, by using proof by contradiction.

You are assuming the premise, and the negation of the consequent.

If we call $P: a\mid b$, and $Q: a\nmid c$, and $R: b\nmid c$,

You've assumed $P\land Q$ and $\lnot R = b\mid c$

$P\land Q\tag 1$

$P\;\;\tag{from (1)}$

$Q\;\;\tag {from(1)}$

$\quad|\lnot R\tag{(2): Assumption }$

$\qquad||\lnot Q\;\;\tag{as given by asker}$

$\qquad|| \lnot P \lor \lnot Q\tag{disjunction intro}$

$\qquad|| \lnot (P \land Q)\tag {DeMorgan's}$

$\qquad|| (P\land Q) \land \lnot (P\land Q)\tag{conjunction intro}$

$\lnot (\lnot R)\tag {follows from contradiction}$

$R\tag{double negation}$

So we have proven $$\big((P\land Q)\land \lnot R\big) \to \lnot(P\land Q)$$

We've reached a contradition. We conclude $(P\land Q) \to R$.

But note, we can are essentially done when we arrive at $\lnot Q$, because we reach $Q\land \lnot Q \equiv \bot$, which is where you stopped (having obtained a contradiction). Logically, we have proven $\lnot\lnot R$, or $R$, and hence $(P\land Q) \to R$.

amWhy
  • 209,954
  • I hope there is a small typo, in defining Q. – jitender Feb 15 '18 at 22:57
  • jitender: Yes, sorry! fixed. – amWhy Feb 15 '18 at 22:59
  • Sorry for late response. Request to state the details how $\lnot(\lnot R)$ is arrived at. I feel that it should be $R \wedge \lnot R$, as have got $\bot$ configuration. Also, I am not able to understand why have used $\mid \mid$ symbols in lines from the mid-half. – jitender Feb 27 '18 at 01:49
  • In fact, it seemed to me that the single vertical line is for assumptions derived from what is given; while the double vertical line is for derivations arrived at. But, an answer ( https://math.stackexchange.com/a/91673/513178 ) at MSE has shown it for just steps numbering, and hence a symbolic convenience. I have googled for books, sites using many combinations of the search terms : " logic, |, ||, ¬("for proofs done by arriving at contradiction"), implication"; but got nothing except the given MSE answer. – jitender Feb 27 '18 at 07:44
  • I would be highly thankful if could please tell how the conjunction of $(P\wedge Q) $, $\lnot(P\wedge Q)$ is done. I think it should be disjunction that is ONLY possible, if take equalities (i), (ii) & form their conjunction: (i) $P\wedge Q \implies R \equiv$$\lnot P \cup \lnot Q \cup R$, & its contra-positive, i.e., (ii) $\lnot R \implies \lnot(P \wedge Q) \equiv$$ \lnot R \cup \lnot P \cup \lnot Q \equiv \lnot P \cup \lnot Q \cup \lnot R$. On forming conjunction, leads to: $ (\lnot P \cup \lnot Q \cup R)( \lnot P \cup \lnot Q \cup \lnot R)$. This not leads anywhere, please help. – jitender Feb 27 '18 at 11:17
  • I request clarification on the above three points.(a) Why $\lnot(\lnot R)$ is taken, instead of the $\bot$ form of $R \wedge \lnot R$. (b) Why two vertical sticks/markers are used, in the later half. (c) How the conjunction is possible even, for the two propositions : $(P\wedge Q), \lnot(P \wedge Q)$. My queries are in final stage, and no further addition or modification is seemingly possible to me. – jitender Feb 27 '18 at 12:10