# Three-statement biconditional [closed]

If you want to show that $$p\iff q\iff r,$$ do you then have to show that $$p\iff q , p\iff r , q\iff r ?$$

• You can omit "$p\iff r$" because of the transitivity of "iff". You can also show $p\implies q$ , $q\implies r$ and $r\implies p$ Sep 20 at 6:06
• Are you using the $\Leftrightarrow$ as an expression of equivalence (in which case we typically interpret an expression like this as saying that $p$, $q$, and $r$ are all equivalent to each other, or do you use as the logical connective called the material biconditional? (more typically written as $p \leftrightarrow q \leftrightarrow r$)? Because if the latter, then you are dealing with a single logic expression that would be true if and only if either $p$, $q$, and $r$ are all true, or if exactly one of them is true. Sep 20 at 13:06
• Re: Peter's comment. Just because Y implies Z does not generally mean that writing a proof of Y is the same as writing a proof of Z, because to complete a proof of Z you want also to at least observe the Y implies Z relationship. Similarly, just because (p⟺q) and (q⟺r) implies (p⟺q) and (q⟺r) and (p⟺r) does not in itself mean that the former and latter require the exact same proof. More to the point: (p⟺q) and (q⟺r) and p⟺q⟺r require the exact same proof because they are literal translations of each other, and neither requires proving p⟺r because neither is asserting that p⟺r! Sep 20 at 13:46
• I thought of a nice example today, where there are four equivalent statements, and we typically show that each of the six equivalences $X\iff Y$ holds not by showing all six, but by showing only that $a\implies b, b\implies c, c\implies d,$ and $d\implies a$.
– MJD
Sep 22 at 13:39
• The example is: If a language is represented by a regular expression, it can be recognized by an NFA with $\epsilon$-transitions. If a language can be recognized by an NFA with $\epsilon$-transitions, it can be recognized by an NFA without $\epsilon$-transitions. If a language can be recognized by an NFA without $\epsilon$-transitions, it can be recognized by a DFA. And if a language can be recognized by a DFA, it can be represented by a regular expression. See regular language for more details.
– MJD
Sep 22 at 13:41

If you want to show that $$p\iff q\iff r,\tag1$$ do you then have to show that $$p\iff q , p\iff r , q\iff r ?$$
I'm replying tangentially to the intent of your question. Here are three reasonable ways to read sentence $$(1),$$ with each main connective coloured red: $$(p\iff q)\color\red\iff r\tag A$$ $$p\color\red\iff (q\iff r)\tag B$$ $$(p\iff q)\color\red{\text{ and }}(q\iff r).\tag C$$
Readings A and B are equivalent to each other. However, neither is equivalent to the common mathematics reading, C; you can see this by supposing that $$p,q,r$$ are all false.
By calling sentence $$(1)$$ a three-statement conditional, you are almost certainly adopting reading C while observing its consequence that p⟺r (here, your three statements refer to the two conjuncts of C and p⟺r); therein lies your answer: to prove that reading C is true does not require proving that the conjunction C itself as well as some consequence of it are true. (After all, surely it is clear that proving C does not require additionally proving its consequence p⟺(q∧r) either.)