# Tag: “Math Improve”

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- ⊢ ((A =⇒ B) =⇒ A) =⇒ A [closed] November 14, 2024This speific statement is to be proved by using deduction theorum and a hilbert style proof. I tried using deduction deduction theorum and proof by contradicton to get (A =⇒ B) =⇒ A),¬A ⊢ ⊥. But Having problem deriving to ge to bottom. Can anyone guide where to go from this. There Are Certain Axioms […]Ali Mehdi Jafri
- Proving that $s \sim t$ iff $S\vdash s = t$ is an equivalence relation. November 13, 2024As part of a proof of Gödel's Completeness Theorem, we have to show that $s \sim t$ iff $S\vdash s = t$ is an equivalence relation. I was wondering if this could be done by natural deduction using the rules $\def\tI{\text{intro}} \def\tE{\text{elim}}$ $$\frac{}{( t = t)}(=\tI) \ \ \ \ \text{ and } \ \ […]Sam
- Statements equivalent to axiom of infinity November 13, 2024In Zermelo-Fraenkel set theory, one of the axioms states: $\exists I(\emptyset\in I \land \forall x(x\in I \Rightarrow (x\cup\{x\})\in I)$ $I$ can ve called an inductive set. Having such set, we could choose its minimal subset, which satisfy the condition (by getting an intersection of all inductice subsets of a given one). We could also prove […]kostya2139
- Trouble understanding the proof of the Second $\Sigma_3$ Representation Theorem November 12, 2024The Second $\Sigma_3$ representation theorem states that if $A$ is $\Sigma_3$ then there is a computable function $h(x,y)$ such that when $x \in A$ there is a unique $W_{h(x,y)}$ which is infinite (in which case it is $\omega$) and when $x \not \in A$ then each $W_{h(x,y)}$ is finite. I am having a lot of […]rea_burn42
- Is pulling out existential quantifier from an implication constructively valid? November 12, 2024I was unable to prove the following "obvious" schema of puling out a quantifier constructively: $$(\phi\rightarrow\exists x\psi(x))\rightarrow \exists x(\phi \rightarrow\psi(x))$$ The proof is straight forward if it can be assumed that $\phi \lor\neg \phi$. The proof of the converse is also straight-forward. Background: This property could be used to prove the standard Axiom of Choice […]Magemathician
- Recursive Definition of Var(φ) and Kon(φ) in Logical Formulas – Should I Use Parsing Trees? November 12, 2024I'm working on a problem in my logic course where I need to define two recursive functions: $Var(\phi)$: Counts the number of variable occurrences in a logical formula $\phi$. $Kon (\phi)$: Counts the number of connectives (excluding negations) in $\phi$. For instance, if $\phi = \neg(p \land p) \rightarrow ((\neg r \lor q) \land q)$, […]asfasfasf
- Structural Induction Proof for Recursive Functions on Boolean Expressions November 11, 2024I'm working on a set of problems in my logic course, where we're dealing with recursive functions that count specific types of subexpressions in Boolean expressions. Here’s the context from the previous problems, which define the functions I need to use for the proof in Problem 3. Some background We defined a recursive function called […]asfasfasf
- in Kunen set theory forcing part, how to show this set is function? November 11, 2024I read Kunen's set theory pp.192-3. in this page, example said that p forces certain set is function; ... but I don't understand how to first set can be function; because |P is set of functions, so it seems that there are distinct function x, y in |P s.t. x(0)= 0, y(0)=0, but x contains […]유준상
- Solution verification of proof that $∃x(P (x) → ∀yP (y))$ [duplicate] November 11, 2024The book "How To Prove It - A Structured Approach 3rd Edition" by Daniel J. Vellman contains the following exercise: Prove $∃x(P (x) → ∀yP (y))$. (Note: Assume the universe of discourse is not the empty set.) Please, verify whether my proof is correct. Proof: Suppose $U$ is any particular but arbitrary chosen universe of […]Vlad Mikheenko
- Justifying vacuous truth in a conditional statement by using the concept of vacuous truth in universal quantifiers [duplicate] November 11, 2024$$\require{amssymb} \forall x\in\varnothing; P(x)$$ We know that this is true, because its negation states there is at least one $x$ in empty set so that $\neg P(x)$ is true, which is false; because there isn't anything in empty set! Its negation is false so it is true itself. I was thinking of something like this […]user1471097

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