MHB Is the Equation in a Circle Correct?

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The equation for a circle is confirmed to be correct, with a focus on the properties of disjunction in logic. The discussion highlights that disjunction is idempotent, meaning $x \lor x \iff x$. There is a suggestion to utilize $\LaTeX$ for clearer mathematical representation. Additionally, a note is made about an image being cropped on the right, which may affect clarity. Overall, the conversation emphasizes the importance of proper formatting in mathematical discussions.
hossam killua
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the equation in circle is right ??
 
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Your image seems to be cropped on the right.

For something like this, it would be better to use $\LaTeX$. All of the symbols you need are in the "Set Theory/Logic" section of our "Quick $\LaTeX$" element:

[table="width: 200, class: grid"]
[tr]
[td]Command[/td]
[td]Output[/td]
[/tr]
[tr]
[td]\lor[/td]
[td]$$\lor$$[/td]
[/tr]
[tr]
[td]\And[/td]
[td]$$\And$$[/td]
[/tr]
[tr]
[td]\lnot[/td]
[td]$$\lnot$$[/td]
[/tr]
[tr]
[td]\iff[/td]
[td]$$\iff$$[/td]
[/tr]
[/table]
 
hossam killua said:
the equation in circle is right ??
Yes. Disjunction is idempotent: $x\lor x\iff x$.
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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