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If $(A \rightarrow B) \ \wedge \ (B \rightarrow C) \ \wedge \ (C \rightarrow A)$ does $(A \leftrightarrow B) \ \wedge \ (B\leftrightarrow C)$ hold? And if so, what is this called?

Edit:Fixed the logical symbols and made it more clear

Bram28
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DeadPixel
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3 Answers3

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Because the implication is transitive (syllogism rule), then (for instance), $B\implies C\implies A$, so $B\implies A$ and finally $B\iff A$. I don't know the name of this rule, maybe it could be also called syllogism. However, it is used in many proofs of the series of equivalent conditions.

szw1710
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Yes, this is true. I've usually seen this called "circular implication", though it is a bit unfortunate that sounds like "circular reasoning". A similar wording with less baggage is "cyclic implication". I give a machine-checked formal proof of a (generalization of) this statement here: https://math.stackexchange.com/a/2350102/305738

Derek Elkins left SE
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Yes. You can use the Consensus Theorem:

Consensus Theorem

$(P \lor Q) \land (\neg Q \lor R) \Leftrightarrow (P \lor Q) \land (\neg Q \lor R) \land (P \lor R)$

Applied to what you have:

$(A \rightarrow B) \land (B \rightarrow C) \land (C \rightarrow A) \Leftrightarrow \text{ (Implication)}$

$(\neg A \lor B) \land (\neg B \lor C) \land (\neg C \lor A) \Leftrightarrow \text{ ( Consensus x 3)}$

$(\neg A \lor B) \land (\neg B \lor C) \land (\neg C \lor A) \land (\neg A \lor C) \land (\neg B \lor A) \land (\neg C \lor B) \Leftrightarrow \text{ (Implication)}$

$(A \rightarrow B) \land (B \rightarrow C) \land (C \rightarrow A) \land (A \rightarrow C) \land (B \rightarrow A) \land (C \rightarrow B) \Leftrightarrow \text{ (Equivalence)}$

$(A \leftrightarrow B) \land (B \leftrightarrow C) \land (C \leftrightarrow A)$

Bram28
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