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$\newcommand{\Ra}{\Rightarrow} \newcommand{\Q}{\mathbb{Q}}$

We prove $P \Ra R$ by assuming $P$ then showing some middle $Q$, i. e., $P \Ra Q \Ra R$. What if we have to show $A \Ra B \Ra C$. Do we prove this one implication at a time, say, Assume $A$ then show $A \Ra B$, then $B\Ra C$. Or, do we assume $A, A\Ra B$ which is then essentially assume $A, B$ and then show $C$?

This question is inspired by this one. Let $b\neq 0$. We want to show this by contrapositive. $$ a \not\in \Q \wedge b \in \Q \Ra ab \not\in \Q $$ Then the contrapositive, $$ ab \in \Q \Ra a \in \Q \vee b \not\in \Q $$ Is the same as, $$ ab \in \Q \Ra b \in \Q \Ra a \in \Q $$ The answer there assumes $ab \in \Q$. Which makes sense, but then they moreover assume $b\in \Q$. Why? Should there not be some intermediate steps showing that $$ ab \in \Q \Ra ? \Ra b \in Q. $$ What am I missing here?

scribe
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  • As the king told Alice, you begin at the beginning, and go on till you come to the end: then stop. – Mariano Suárez-Álvarez Feb 01 '23 at 02:46
  • Less literally, "We prove P⇒R by assuming P then showing some middle Q, i. e., P⇒Q⇒R" does not seem to make much sense. What is Q? One generally proves an implication such as P⇒R by assuming P and then proving R. – Mariano Suárez-Álvarez Feb 01 '23 at 02:48
  • If Alice wants to prove to Bob that $P\Rightarrow R$ and Bob already agrees that $Q\Rightarrow R$ then Alice shows Bob $P\Rightarrow Q \Rightarrow R$. – scribe Feb 01 '23 at 02:53
  • How would one guess that from what you wrote int he question? – Mariano Suárez-Álvarez Feb 01 '23 at 02:53
  • Also, an expression such as A⇒B⇒C is in almost all cases understood to mean the conjunction of A⇒B and B⇒C. If you want to write something like A⇒(B⇒C) or (A⇒B)⇒C you have to explicitly write the parentheses (the logical operator ⇒ is not associative) – Mariano Suárez-Álvarez Feb 01 '23 at 02:56
  • Well, I was somewhat confused about the partial (right) associativity of the implication but I did try to say this via "We prove $P \Rightarrow R$ by assuming $P$ then showing some middle $Q$"... Thanks for pointing it out. – scribe Feb 01 '23 at 03:00
  • If Alice wants to tell Bob that A⇒B⇒C and Bob knows that B⇒C then what Alice has to do is to show him that A⇒B, not that A⇒B⇒C (which, as I said, is really bad notation) – Mariano Suárez-Álvarez Feb 01 '23 at 03:01
  • "One generally proves an implication such as $P\Rightarrow R$ by assuming $P$ and then proving $R$." By $Q$, I meant, it is something that is "proving $R$" having assumed $P$. – scribe Feb 01 '23 at 03:07
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    In case you require further clarification: multiple equivalences and another example – ryang Feb 06 '23 at 05:54

1 Answers1

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Suppose I have three statements $A,B,C$. Then, the following are all logically equivalent (just write out a simple truth table for them all or use whatever rules of logic you know): \begin{align} [A\implies (B\text{ or } \neg C)]\equiv [A\implies (C\implies B)]\equiv [(A \text{ and } C)\implies B]. \end{align} For your question specifically,

  • $A$ is the statement that $ab\in \Bbb{Q}$,
  • $B$ is the statement that $a\in\Bbb{Q}$,
  • $C$ is the statement that $b\in\Bbb{Q}$.

Even if you didn't know about the logical equivalences, the following should make sense intuitively. Let's say after writing the contrapositive, we start by assuming $ab\in\Bbb{Q}$. Ok, now we have two cases:

  • Case 1: $b\notin\Bbb{Q}$. In this case, the 'or statement' "$a\in\Bbb{Q}$ or $b\notin\Bbb{Q}$" is trivially satisfied.
  • Case 2: $b\in\Bbb{Q}$. After some work, you'll end up showing $a\in\Bbb{Q}$, so again, the "or" statement is satisfied.

So, in either case, the "or" statement is satisfied, which completes the proof of the whole implication.

peek-a-boo
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