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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 $x \in (A \cup B \cup C)$

- Decomposition by subtraction
- Count the number of topological sorts for poset (A|)?
- Stirling numbers of the second kind on Multiset
- Prove that if $2^n - 1$ is prime, then $n$ is prime for $n$ being a natural number
- About the Stirling number of the second kind
- Confused by proof of the irrationality of root 2: if $p^2$ is divisible by $2$, then so is $p$.

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 $A \cup B \cup C$?

- Natural Numbers and Well ordering
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- The union of a sequence of countable sets is countable.
- Proving the sum of the first $n$ natural numbers by induction
- What is bigger, $p(\mathbb{N})$ or $\mathbb{R}$?
- Every planar graph has a vertex of degree at most 5.

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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