query: Prove $(s \to p) \to (s \to q)$ using propositional logic
positive_passages.docid: 147177
positive_passages.text: Swapping implications - I have no idea where to begin with this
negative_passages.docid: 49385
negative_passages.text: Proving $\forall x (A\to B) \to(A \to \forall x B):x\notin \mbox{free}(A)$ in a Hilbert system where it is not an axiom
negative_passages.docid: 103653
negative_passages.text: Interpretations and models in propositional logic
negative_passages.docid: 48606
negative_passages.text: Help with a Predicate Logic homework proof
negative_passages.docid: 34768
negative_passages.text: Proving $(p \oplus q) \oplus r=p \oplus (q \oplus r)$
negative_passages.docid: 52095
negative_passages.text: Prove either $G=ST$ or |$G|\geq|S|+|T|$
negative_passages.docid: 113605
negative_passages.text: In Fitch, how does one prove "(P → Q)" from the premise "(¬P ∨ Q)"?
negative_passages.docid: 52585
negative_passages.text: How to prove that $[(p \to q) \land (q \to r)] \to (p \to r)$ is a tautology without using the truth table?
negative_passages.docid: 106401
negative_passages.text: Prove: $p-mq | f(m)$ where $m$ is any integer
negative_passages.docid: 75381
negative_passages.text: How can we reason about "if P then Q" or "P only if Q" statements in propositional logic?
negative_passages.docid: 72034
negative_passages.text: Truth table and the meaning of $\oplus$ in propositional logic
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: 109778
negative_passages.text: Proving $(x-c)|(p(x)-p(c))$
negative_passages.docid: 6119
negative_passages.text: If both $P$ and $Q$ are true , how can I tell that $P$ implies $Q$?
negative_passages.docid: 29110
negative_passages.text: Propositional Intuitionistic Logic Question
negative_passages.docid: 75007
negative_passages.text: How can I prove this relation between the elementary set theory and the elementary logic?
negative_passages.docid: 67709
negative_passages.text: Stuck with proof for $\forall A\forall B(\mathcal{P}(A)\cup\mathcal{P}(B)=\mathcal{P}(A\cup B)\rightarrow A\subseteq B \vee B\subseteq A)$