**Written work: ***NOTE: The following suggested problems are for practice only, and will NOT be collected.*

*Section 11.4 p194: 2, 3, 5, 6, 7*

*Handout: Theorem NT 6.2, 6.3*

**WeBWorK – **none

**OpenLab** – none

**Project Reflection** – Due before the final exam, Thursday 12/20.

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### Logic on Math StackExchange

- Proving sentence using resolution October 6, 2022I want to prove the following tautology using resolution: $$\exists x \forall y P(y,x) \to \forall x \exists y P(x,y)$$ The negation of the sentence in Skolem normal form is $$\forall y \forall v(P(y,c) \land \neg P(f(y),v))$$ for some function symbol $f$ and constant symbol $c$ and variables $x,y,u,v$ but this seems not unifiable since […]Rudinberry
- How do I prove FORMALLY that $\forall xPx\land\forall xQx\iff\forall x(Px\land Qx)$? October 5, 2022I already understand the statement intuitively, so don't try to give me an intuitive explanation as that will not be helpful. I would certainly be willing to "give my own attempts at solving the problem", but I legitimately have no idea on how to go about it. These "rules" (they're called "rules of inference" I […]Dark Rebellion
- Asking whether $\infty$ is an integer October 5, 2022Consider the extended reals, $\mathbb{R} \cup \{\infty\}$. Suppose we ask the question of whether $\infty$ is an integer. It isn't an integer - that is, it is some non-integral value. But then, by the Archimedean property, there exists an integer larger than it, which is contradictory. It is an integer. But again, by the Archimedean […]Aryan Dugar
- How are axiom schemata of ZF defined (without using sets)? [duplicate] October 5, 2022ZF is defined using axiom schemata, rather than a finite set of axioms. So ZF has an infinite (countable) set of axioms. I realized that in my study of math I probably missed how matching a statement to be an instance of an axiom schema is defined. We can't refer to the fact that schemata […]porton
- Does every non-contradictory formal system have a model in ZF? October 5, 2022Is the following statement true? Every non-contradictory formal system has a model in ZF. This seems true, because non-contradictness and existence of model seems equivalent. But how to prove this?porton
- Equivalence between the set arithmetic's and logical connectives October 4, 2022I would like to know where the equivalence between the set arithmetic's (i.e. intersection, inclusion...) and the logical connectives (i.e. and, or, not...) comes from. Is it from definition, or a consequence of certain properties?tristan ledet ledet
- Formal reason why proof by arbitrary element works. October 4, 2022When we prove $\forall x \in D, P(x),$ we often do it so by taking an arbitrary element of $D,$ and proving for it $P(x).$ I was thinking for more of a formal reason why the arbitrary method works. I know the verbal reason why arbitrary works: arbitrary element is a generic element of the […]medium_o
- How to prove Drinker paradox using Resolution? October 4, 2022There exists a famous Drinker paradox: $\exists x[P(x)\implies \forall yP(y)]$ How to prove it using resolution? Here is my attempt: Negated and Skolemized: $\forall x$ $P(x) \wedge \neg P(f(x))$ Clausal Form: {P(x)}, {~P(f(x))} Stuck at this step. I do not understand how to unify these terms to do resolution and produce an empty clause.Oleg Dats
- Infimums of subsets of monoid. October 4, 2022I came up with the following idea This may have already been studied by someone else or may be a common sense idea in this field. Let $(M,+,\leq)$ be a totally ordered commutative monoid. For any non-empty subsets $A,B$ of $M$, we define a relation $A\precapprox B$ to be that for any $b\in B$, there […]M masa
- Peano arithmetic, definitional equivalence and mutual interpretability October 4, 2022I have read in John Corcoran's "Mutual Interpretability is not Definitional Equivalence" (meeting abstract, 1979, p. 430) that the following two "second-order" axiomatizations of Peano arithmetic (I will call them $T$ and $T'$) are mutually interpretable, but not definitionally equivalent. I don't see why. Here are the two formulations ('s' is the successor functor). See […]Soennecken

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