How would one prove that the golden rule is a tautology without a truth table?

$A\,\land\,B\,\equiv\,A\,\equiv\,B\,\equiv\,B\,\lor\,A $

I've been searching a lot and cannot find a demonstration. I've tried to simplify the formula using another inference rules to reach a true value, without success.

Any help or at least a hint would be very appreciated.

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    $\begingroup$ parentheses, please $\endgroup$ – spaceisdarkgreen Dec 12 '17 at 4:00
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    $\begingroup$ @rschwieb not sure it's OP's fault.... see pg 7 here: cs.nott.ac.uk/~psarb2/G51MPC/slides/LogicalConnectives.pdf $\endgroup$ – spaceisdarkgreen Dec 12 '17 at 4:03
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    $\begingroup$ @spaceisdarkgreen yeah... what a ridiculous notation. $\endgroup$ – rschwieb Dec 12 '17 at 4:06
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    $\begingroup$ Just to help understand why people are clamoring for you to clarify your notation, normally, this form of "chaining" equivalences is meant to be read as shorthand for $$(A \wedge B \equiv A) \wedge (A \equiv B) \wedge (B \equiv B \vee A)$$ which, presumably, isn't anything like what you mean by the formula. $\endgroup$ – Hurkyl Dec 12 '17 at 4:06
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    $\begingroup$ @spaceisdarkgreen: The $\leftrightarrow$ connective is actually associative and commutative in classical logic, so each of the bracketings should be equally provable, $\endgroup$ – Henning Makholm Dec 12 '17 at 4:25

$$(A \land B) \equiv A \equiv B \equiv (B \lor A) \Leftrightarrow$$

$$((A \land B \land A) \lor (\neg (A \land B) \land \neg A)) \equiv (((B \land (B \lor A)) \lor (\neg B \land \neg (B \lor A))) \Leftrightarrow $$

$$((A \land B) \lor ((\neg A \lor \neg B) \land \neg A)) \equiv (B \lor (\neg B \land \neg B \land \neg A)) \Leftrightarrow$$

$$((A \land B) \lor \neg A) \equiv (B \lor (\neg B \land \neg A)) \Leftrightarrow $$

$$(B \lor \neg A) \equiv (B \lor \neg A) \Leftrightarrow$$


I used the following equivalences:


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


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

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


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

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

Biconditional Tautology

$P \equiv P \Leftrightarrow \top$

  • $\begingroup$ Thanks man, this is what i was looking for. $\endgroup$ – Mariano Córdoba Dec 12 '17 at 5:57

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