Logical Equivalence Puns

How would you prove that without a truth table

I'm told that Prolog without cut or any other meta-logical predicate is not Turing equivalent.

Can anyone help me learn some details on this? How is the above fact proven? Is there a simple example of a problem that cannot be solved in Prolog without meta-logical predicates?

Somebody, please help me with trying to prove the logical equivalence:

(s -> -p) V [-(s ^ t)] is logically equivalent to p -> -(s ^ t)

Any of the laws can be used.

I would really appreciate it if you list down the steps, thank you.

I want to fight off people who use argumentation and "knowledge" for no good (sexism, racism, transphobia and dictatorship and ellitism and etc.), for that I want to suggest a replacement for "logical thinking" (i.e. typical argumentation) -- "equivalence thinking".

I want to give you a really new idea that's valuable regardless of your (dis)agreement. It may be not new, garbage or too ambitious if I'm wrong though.

But it's also my honest opinion: I genuinely want this philosophy to change the world for the better, my loving one lives in an oppressed autocratic protesting european country...

Let me explain my ideas/suggestions:

1st insight/idea:

Imagine you can treat an opinion as just a bag of separate concepts without (much) structure (like bag-of-words). And somehow evaluate those separate concepts "directly" and decide if the opinion is true or false based on that...

OK. But how can you possibly do such "direct" evaluation?

2nd insight/idea:

You don't need logical connection between concepts if you only have 1 concept.

So, the first thing you can do is check if you can merge all the concepts in an opinion into a single one.

With some specifications, this already gives us Kant's Categorical imperative: Categorical imperative takes in a bunch of concepts, "universalizes" them and checks if they can be merged or if they can't (the latter leads to contradiction)... for example, the idea of universilized "theft" doesn't make sense because it presupposes both "personal property" and its negation (contradictory non-mergeable concepts)

3rd insight/idea:

You can choose a single concept -- your most important value.

It's something you value in of itself, unconditionally, not as a mean to a goal, something you can't quantify 'cause there's no reason to do so (you don't need to accumulate or trade it).

You then can check if the concepts in an opinion are equivalent to that "most important" concept. If YES the opinion is true, if NO the opinion is false... (unless you get rid of unwanted concepts)

This is the simplest type of thinking/evaluation of its kind: what can be more simple than a single concept?

This is "equivalence thinking". Logical thinking can be defined as deviations from the "equivalence thinking".

**Some quest

I’m really stuck in this question at 3 am and I just wanna sleep.

Do we need to check logical equivalence between RTL design and post synthesis netlist or post-implementation netlist while we are doing FPGA designs? If the answer is yes, do we need additional 3rd party tools except from vendor tools?

So i have to show that (p -> q) v ( not p -> q ) is logically equivalent to ( p ^ q ) -> ( not p <-> q) with a truth table. I tried doing it, but there are some differences in the 2 truth table I got. Can i say that its logically equivalent if it is mostly the same? Would love some help please.

I think some concrete examples would really help.

https://preview.redd.it/47rcni3t8eg41.png?width=431&format=png&auto=webp&s=e3bdad2ad7c4d383bbfbfc9b6cdcc285d272ec25

I know if the ORs are switched to an AND, it will be logically equivalent but i need help on OR.

So we were discussing the above lesson and came upon the following statement: Prove: ~[Pv(~P ^ Q)] = ~(PvQ) This was to be solved 2 ways, using the truth table and the laws of logical equivalence. Solving via truth table found the equation to be true, but we were not able to prove this with the laws. Can anyone please help confirm if either the statement is correct, the truth table is correct, or the laws required to prove said statement? Thanks in advance lol

For logical equivalences I understand this is to show for two statements—p and q—it is to show p ↔ q. I was hoping to seek some clarity on whether or not these two statements are logical equivalent.

"There is no gravity on Earth."

"Humans are not held down by gravity."

I came to a disagreement with someone about whether or not these two statements were logically equivalent. It essentially boiled down to having to consider the definitions of the words and what they fundamentally meant. This did not sit right with me as it felt as if we were using information outside of the question. Any help is appreciated!

If this isn't the correct avenue to post a question like this, please let me know.

In a rule to a logic game: "If F is on A, M will be on B," how is Ma-> Fb a contrapositive to Fa -> Mb? I thought the logical equivalence of Fa -> Mb is Negative Mb -> Negative Fa.

I'm really struggling with this.

hello, my question is how to write p->q(~q^r) using only ~ and ^ as connectives.

I was able to get it to ~pV(~q^r) but I don't know where to go from there

Am I right in saying that "→" is logically equivalent to "⊢" and "⊨"? It seems like the syntactic and semantic entailments can be viewed as kinds of "→"?

Sorry for the basic question, I'm working from a PDF of exercises right now, and it doesn't include descriptions.

Hi, I'm trying to follow the laws as instructed however I often find myself stuck because my current proposition shows that no law can further be applied to it. e.g the two propositions are (p ^ q) V - (p > - r) = p ^ (q V r)

• step 1. (p ^ q) V - (- p V - r) = conditional statement
• step 2. (p ^ q) V (p V - r) = double negation
• step 3. ?

There are no laws that allow me to proceed to step 3! I know I have to get rid of (p ^ q) V (p V - r) but I don't know how to if I'm exactly following the laws symbol for symbol!

Please help me!

Select all the sentences that are logically equivalent to 𝑥>1⇒𝑦<−1

A. 𝑥≤1⇒𝑦≥−1

B. 𝑦<−1⇒𝑥>1

C. 𝑦≥−1⇒𝑥≤1

D. 𝑦≥−1 or 𝑥>1

E. 𝑥≤1 or 𝑦<−1

F. None of the above

I need to show that (not Q) → (R → not(P and Q)) is equivalent with (Q or(not Q))

Here's what I've done so far: (not Q) → (R → not(P and Q))

• not(not Q) or (not R or not(P and Q)) (Rewriting P→Q is the same as (not P) or Q)
• Q or (not R or not(P and Q)) (Double negation)
• Q or (not R or ((not P) or (not Q))) (DeMorgan's Law)
• Q or (((not R) or (not P)) or (not Q)) (assosiative or)
• Q or (not(R and P) or (not Q)) (DeMorgan's Law)

And this is where I'm stuck. I'm not sure what to do next here because I cant find any laws that can help me continue.

Any answer is much appriciated!

Edit: Formatting