query: Why is $P \rightarrow Q$ equivalent to $\lnot P \vee Q$ in propositional logic CNF?
positive_passages.docid: 578
positive_passages.text: In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False?
negative_passages.docid: 111862
negative_passages.text: Proof for sequent (Logic)
negative_passages.docid: 48368
negative_passages.text: Prove using Natural Deduction in predicate logic $(\forall x\ P(x))\to Q ⊢ \exists x (P(x)\to Q)$
negative_passages.docid: 32027
negative_passages.text: Given (p ∧ q), use the Fitch system to prove (q ∨ r)
negative_passages.docid: 91947
negative_passages.text: Proof of $ f:I_n \rightarrow I_m \Rightarrow n = m$
negative_passages.docid: 31299
negative_passages.text: Using the Fitch system, how do I prove $((p \implies q) \implies p)\implies p$?
negative_passages.docid: 65379
negative_passages.text: Deriving $\neg R$ from $\{R↔(R∨(P∧¬P)), R↔¬P, ¬P→(P↔(Q→Q)), P→Q\}$
negative_passages.docid: 35236
negative_passages.text: Proof $p \vdash q \Rightarrow p$
negative_passages.docid: 87928
negative_passages.text: Proof of logical equivalence of biconditional and other proposition
negative_passages.docid: 85770
negative_passages.text: What is the difference between ∀x(p(x) →q(x)) and ∀x(p(x) ∧q(x))?
negative_passages.docid: 93410
negative_passages.text: How to prove ~ (~P & ~Q) : P ∨ Q by natural deduction
negative_passages.docid: 140759
negative_passages.text: What is exactly the difference between $\forall x \neg P(x)$ and $\neg \forall xP(x)$?
negative_passages.docid: 98386
negative_passages.text: First-order logic and second-order logic confusion
negative_passages.docid: 46613
negative_passages.text: What's the difference between $\equiv$ and $\leftrightarrow$ in a formal proof?
negative_passages.docid: 103653
negative_passages.text: Interpretations and models in propositional logic
negative_passages.docid: 9282
negative_passages.text: Prove using natural deduction $((P \land Q) \rightarrow R) \vdash (P \rightarrow R) \lor (Q\rightarrow R)$
negative_passages.docid: 115004
negative_passages.text: If $x\notin\mathbb Q$, then $\left|x-\frac{p}{q}\right|<\frac{1}{q^2}$ for infinitely many $\frac{p}{q}$?