Week 2 Assignments
Written work – none
WeBWorK – Assignment #1 and #2, due Tuesday, September 6th, at midnight.
OpenLab – OpenLab #2, due Thursday, September 8th, at the start of class.
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- Is this combination of Lukasiewicz proof systems complete? May 29, 2024I’ve been learning more about axiomatic systems, and somewhat recently came across the following axiomatization for Classical Propositional Calculus a la Lukasiewicz: $(A \to B) \to (B \to C) \to A \to C$ $A \to \neg A \to B$ $(\neg A \to A) \to A$. Along the way, I’ve learned that proving $(A \to (A […]PW_246
- Relation between existential and universal quantifiers. May 29, 2024Here is the proof of the theorem $ \neg (\forall x) A(x) \equiv (\exists x)\neg A(x) $ Proof: Let the universe of discourse be $U$. The sentence $ \neg (\forall x) A(x)$ is true in $U$ $\implies (\forall x) A(x)$ is false in $U$ $\implies$ the truth set of $ A(x)$ is not the universe […]Mystic mystic
- How do you prove the theorems in Gödel's ontological proof? May 29, 2024I will be referencing the proof that can be found here: Gödel's ontological proof - Wikipedia In Gödel's ontological proof there are axioms and definitions, which of course do not need to be proven; however, four theorems are stated, and I could not find proof for any of them by searching the internet. I was […]Elvis
- Why is exclusive disjunction not a standard logical operation? May 29, 2024This question Rules of inference for exclusive disjunction and logical biconditional shows that the inference rule for exclusive disjunction are little different from the other rules for conjunction, disjunction, implication, etc. Then why is the exclusive disjunction so often neglected/left out? I know it can be defined a posteriori without having it as a primary […]richardIII
- Choosing between the 'and' and 'implies' connectives May 29, 2024$pol(x): x$ is a politician $liar(x): x $ is a liar All politicians are liars : $\forall x(pol(x) → liar(x))$ Some politicians are liars : $\exists x(pol(x) \land liar(x))$ No politicians are liars : $\forall x(pol(x) → ¬liar(x))$ Some politicians are not liars : $\exists x(pol(x) \land ¬liar(x))$ So, do we just use 'and' for […]user1327299
- Formalities on loop invariant - algorithms May 28, 2024When proving an algorithm using a loop invariant, we need to check these three things. The loop invariant holds before the loop is entered (initialization) If the loop invariant holds before the loop is entered then the loop invariant holds before entering the next iteration of the loop (maintenance) The loop is finite We can […]Agustin G.
- I literally cannot understand the logic in tableaux, can someone guide me to really understand what I have to do here? [closed] May 28, 2024(¬H)∨H,H→H,H↔H,H↔(H∧H),H↔(H∨H),H↔(H∧(H∨G)),H↔(H∨(H∧G))Guilherme Mourão
- Is it possible to invoke the empty function? May 28, 2024The empty function $f: \emptyset \rightarrow X$ is a function from the empty set to an arbitrary set $X$. Since $\emptyset$ is the only subset of $\emptyset \times X = \emptyset$, $f$ must be equal to $\emptyset$. Is it possible to invoke the empty function? To clarify what I mean, consider a function $g: \{ […]Vlad Mikheenko
- How to prove that two algebraically closed field with same characteristic and cardinal are isomorphic? [duplicate] May 28, 2024Hi I would like to prove that for algebraic closed field $F_1,F_2$ if: They have the same characteristic $p$ They have the same cardinal and is uncountable Then they are isomorphic. I have proved that with the same characteristic $F_1,F_2$ has isomorphic sub-field, donate $F'$, and $F_1,F_2$ are vector space over $F'$ I think the […]Shore
- How to prove expended fields whose base has same cardinal are isomorphic? May 27, 2024I would like to prove that for field $F_1,F_2$ if they : have isomorphic sub-field $F'$ $F_1,F_2$ is the vector space over $F'$, and their bases, doneta $B_1,B_2$, has the same cardinal Then $F_1,F_2$ are isomorphism. It is clear that for any $a\in F_1$ we can linearly express it as $a=\sum_{i=1}^nk_ie_i$ where $k_i\in F'$ and […]Shore
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