query: How to prove logic equivalence in Coq?
positive_passages.docid: 157625
positive_passages.text: How to prove conditional in Coq?
negative_passages.docid: 126870
negative_passages.text: Show that axiom of replacement implies from the Axiom of Specification [Proof Verification]
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: 85507
negative_passages.text: Equivalence Relation problem
negative_passages.docid: 93410
negative_passages.text: How to prove ~ (~P & ~Q) : P ∨ Q by natural deduction
negative_passages.docid: 157490
negative_passages.text: Statements equivalent to the Axiom of Choice
negative_passages.docid: 99911
negative_passages.text: How to show that if $x=(p,q)$ and $y=(p',q')$ then $x \sim y $ if $pq'=p'q$ is an equivalence relation?
negative_passages.docid: 79407
negative_passages.text: How to solve ~(P → Q) : P & ~Q by natural deduction?
negative_passages.docid: 62936
negative_passages.text: Proving equivalencies of two versions of the Axiom of Choice
negative_passages.docid: 51369
negative_passages.text: Natural deduction proof of $p \lor (p\implies q)$ with propositional calculus
negative_passages.docid: 120911
negative_passages.text: Clarification on proof: Order of left cosets equal
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: 98630
negative_passages.text: For all $x,y\in\mathbb{R}$ define that $x\equiv y$ if $x^{2}=y^{2} $; prove $\equiv$ is an equivalence relation.
negative_passages.docid: 25187
negative_passages.text: Why $∀x(P(x)→Q)$ is equivalent to $∃xP(x)→Q$
negative_passages.docid: 140682
negative_passages.text: Proving a property of a Logic Formal Language
negative_passages.docid: 111862
negative_passages.text: Proof for sequent (Logic)
negative_passages.docid: 113605
negative_passages.text: In Fitch, how does one prove "(P → Q)" from the premise "(¬P ∨ Q)"?