query: Can you represent other logical operations using only $\neg$ and $\Leftrightarrow$ (not and equivalence)?
positive_passages.docid: 149416
positive_passages.text: Show that $ \{\lnot,\leftrightarrow\} $ is not functional complete
negative_passages.docid: 87955
negative_passages.text: How to prove that $P \rightarrow Q$ is equivalent with $\neg P \lor Q $?
negative_passages.docid: 108028
negative_passages.text: Is there a bijection between the reals and naturals?
negative_passages.docid: 147947
negative_passages.text: Does double negation distribute over implication intuitionistically?
negative_passages.docid: 47678
negative_passages.text: Is it ever bad to use \left and \right?
negative_passages.docid: 3075
negative_passages.text: Is there a rule about double negations that aren't meant as double negations (e.g. "We don't need no education")?
negative_passages.docid: 5293
negative_passages.text: Show that the logic $ \ \forall x \exists y \ P(x,y) \ $ is not logically equivalent to $ \ \exists x \forall y \ P(x,y) \ $
negative_passages.docid: 8342
negative_passages.text: Is short-circuiting logical operators mandated? And evaluation order?
negative_passages.docid: 142526
negative_passages.text: What's the intuition behind difference of $\neg(S\land E)$ and $\neg S \land \neg E$?
negative_passages.docid: 121375
negative_passages.text: Is it safe to temporarily redefine \and, \or, and \not?
negative_passages.docid: 93557
negative_passages.text: Intuitively, why does (B → S) ≡ (¬B ∨ S)?
negative_passages.docid: 22245
negative_passages.text: Show the union of two subrings is generally not a subring
negative_passages.docid: 142815
negative_passages.text: Is Left associativity only for Postfix expression?
negative_passages.docid: 38092
negative_passages.text: Symbol for "defined to be logically equivalent" (:⇔)
negative_passages.docid: 85437
negative_passages.text: Python's equivalent of && (logical-and) in an if-statement
negative_passages.docid: 121361
negative_passages.text: Showing that $\lnot Q \lor (\lnot Q \land R) = \lnot Q$ without a truth table
negative_passages.docid: 123458
negative_passages.text: Prove $( \lnot C \implies \lnot B) \implies (B \implies C)$ without the Deduction Theorem