Intereting Posts

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A question about path-independent integral

I need to show that $(A \cup B) \subseteq (A \cup B \cup C)$

My Work So Far:

What I really need to show is that $x \in (A \cup B)$ implies $x \in (A \cup B \cup C)$

- Pigeonhole principle and finite sequences
- How many positive integer solutions are there to the equality $x_1+x_2+…+x_r= n$?
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- Proving $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

So I translated my sets into their logical expressions

$x \in A \vee x \in B \longrightarrow x \in A \vee x \in B \vee x \in C$

This is where I’m stuck. How do I show membership of $A \cup B$ implies membership of $A \cup B \cup C$?

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We need to show that following holds:

$(A∪B)⊆(A∪B∪C)$

**Proof**: Since by definition the above statement means that $(x∈A∨x∈B) \rightarrow (x∈A∨x∈B∨x∈C)$, we now have a new goal. Since our new goal has the form of a conditional, we start by assuming that the antecedent is true, that is, we assume that $(x∈A∨x∈B)$ is true. Then if $(x∈A∨x∈B)$ is true so is $(x∈A∨x∈B)∨x∈C$ (it is formalized in the natural deduction and known as the **disjunction introduction rule**). This shows that $(x∈A∨x∈B) \rightarrow ((x∈A∨x∈B)∨x∈C)$ and, therefore, that $(A∪B)⊆(A∪B∪C)$ holds as required.

In my personal opinion, is always a good practice not omitting parenthesis when your statement starts getting longer (it avoids ambiguity and makes explicit the unique parsing of a formula).

Let’s take for instance the formula you just stated (parenthesis are now explicit):

$(x∈A∨x∈B) \rightarrow (x∈A∨x∈B∨x∈C)$

this has an easier reading than the one omitting them:

$x∈A∨x∈B \rightarrow x∈A∨x∈B∨x∈C$

doesn’t it?

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