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Is it valid to use the implies symbol ($\implies$) or the if and only if symbol ($\iff$) in mathematical proofs to replace writing assumptions? For example, if I want to prove,

$$\text{If } A \subseteq B, \text{ then } A \cup B = B$$

Can I write my proof as follows?

Let A be a subset of B. $$ x \in A \cup B \\ \iff x \in A \vee x \in B \\ \implies x \in B \vee x \in B (\because x \in A \implies x \in B) \\ \iff x \in B \\ \therefore A \cup B \subseteq B \\ x \in B \\ \implies x \in B \vee x \in A \\ \iff x \in A \cup B \\ \therefore B \subseteq A \cup B \\ \therefore A \cup B = B $$

Krup'a
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gldanoob
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    Sure you can, but why? It doesn't make your proof any "better" (presume the ideas are the same whether you use the symbols or not) - but may make it harder to read. –  Jan 15 '20 at 09:54
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    Actually I had to look into the MathJax source to figure out what $\because$ means. Fortunately the macro name in this case spells out the meaning quite clearly, or else a web search would have been necessary as well. – celtschk Jan 15 '20 at 10:56

2 Answers2

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Can I write my proof as follows

This depends on your teacher/intended-readers, but personally I find what you wrote unpleasant to read and relatively difficult to follow, as well as ambiguous in some places (e.g. the next to last "therefore" symbol applies only to those statements that include and follow the statement "$x \in B$", while the last "therefore" symbol doesn't). Something like the following will probably be a lot more digestible and useful to your intended readers.

To prove $A \cup B = B,$ we will prove (1) $A \cup B \subseteq B$ and (2) $B \subseteq A \cup B.$

Proof of (1): We will show $\;x \in A \cup B \; \implies \; x \in B.\;$ To this end, assume $x \in A \cup B.$ Then $x \in A$ or $x \in B.$ Since $A \subseteq B,$ it follows from $x \in A$ that $x \in B.$ Therefore,

$$(x \in A \;\; \text{or} \;\; x \in B) \; \implies \; (x \in B \;\; \text{or} \;\; x \in B) \; \implies \; x \in B.$$

Proof of (2): We will show $\;x \in B \; \implies \; x \in A \cup B.\;$ To this end, assume $x \in B.$ Then it follows from the logical principle "or introduction" that $\;x \in A$ or $x \in B,\;$ and hence $\; x \in A \cup B.$

Dave L. Renfro
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An alternative to Dave's suggestion:

Let A be a subset of B. Then \begin{align}&x \in A \cup B \\ \implies &x \in A \vee x \in B \\ \implies &x \in B \vee x \in B \quad(\text{by assumption}) \\ \implies &x \in B.\tag1\end{align} and \begin{align}&x \in B \\ \implies &x \in B \vee x \in A \\ \implies &x \in A \cup B.\tag2\end{align} From $(1):$ $$A \cup B \subseteq B.\tag3$$ From $(2):$ $$B \subseteq A \cup B.\tag4$$ From $(3)$ and $(4):$ $$A \cup B = B.$$

P.S. I changed all the $\iff$ to $\implies,$ because in this proof the former are not just superfluous but also make its flow jerky.

ryang
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