MHB Venn Diagram: p v (q ^ r) = (p v q) ^ (p v r)

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The discussion focuses on demonstrating the equivalence of the logical expressions p v (q ^ r) and (p v q) ^ (p v r) using a Venn diagram. Participants are encouraged to visualize the union and intersection of sets to understand the relationships between the variables. Additionally, it is highlighted that the expression p v (q ^ r) is not equivalent to (p v q) ^ r, prompting users to explore this distinction independently. The use of resources like Wolfram|Alpha is recommended for further exploration and clarification of these concepts. Understanding these logical relationships is essential for grasping foundational principles in set theory and logic.
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can someone give me show me a venn diagram that will satisfy this statement
Venn diagram to show that the statement p v (q ^ r) is equivalent to (p v q) ^ (p v r) and show that this statement is not equivalent to (p v q) ^ r.
 
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Here is the first half of your request:

Try to do the second one on your own-remember that the "vee" symbol represents UNION, and the "wedge" symbol represents INTERSECTION (when doing a Venn diagram):

View attachment 5083

Wolfram|Alpha (Wolfram|Alpha: Computational Knowledge Engine) is a great resource for budding mathematicians/students, and even just the interested lay-person.
 

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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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