query: Is $p\wedge q\rightarrow p$ a valid argument?
positive_passages.docid: 150053
positive_passages.text: Is "$(P \land Q) \to P$" a tautology?
negative_passages.docid: 91861
negative_passages.text: Propositional Logic: p → q, ¬p → r, ¬q → ¬r ⊢ q
negative_passages.docid: 42150
negative_passages.text: Prove that $\forall x \neg P(x) \iff \neg \exists x P(x)$
negative_passages.docid: 56878
negative_passages.text: Proof that $a\mid b \land b\mid c \Rightarrow a\mid c $
negative_passages.docid: 61962
negative_passages.text: Show $R$ is right-Artinian but not left-Artinian
negative_passages.docid: 88104
negative_passages.text: Prove "$(p \lor q) \land (\neg p \lor r)) \rightarrow (q \lor r)$" is a tautology
negative_passages.docid: 79407
negative_passages.text: How to solve ~(P → Q) : P & ~Q by natural deduction?
negative_passages.docid: 51369
negative_passages.text: Natural deduction proof of $p \lor (p\implies q)$ with propositional calculus
negative_passages.docid: 116527
negative_passages.text: Validating a proof for necessary and sufficient conditions on sets A & B such that $P(A) \cup P(B) = P(A\cup B)$
negative_passages.docid: 32325
negative_passages.text: Can there be a valid argument which has a tautology as a conclusion?
negative_passages.docid: 578
negative_passages.text: In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False?
negative_passages.docid: 105815
negative_passages.text: Does proving (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice?
negative_passages.docid: 124424
negative_passages.text: How to show that there is a deduction of $\varphi\leftrightarrow\psi$
negative_passages.docid: 113605
negative_passages.text: In Fitch, how does one prove "(P → Q)" from the premise "(¬P ∨ Q)"?
negative_passages.docid: 29655
negative_passages.text: $L^p$ and $L^q$ space inclusion
negative_passages.docid: 83407
negative_passages.text: How to show that that $q \mid p-1$?
negative_passages.docid: 41044
negative_passages.text: Equivalence of $a \rightarrow b$ and $\lnot a \vee b$