How to write the proof for $\forall x \in D \, (A(x) \to B(x))$ in standard English.

Question

Question: How to write the proof for $\forall x \in D \, (A(x) \to B(x))$ in standard English.

I recently started reading a book that begins with a short introduction to induction. Although I have a lot of experience writing proofs using induction, I mainly write these proofs in another language. Therefore, I am slightly confused with the specific word choice in English. When trying to phrase for a proof that shows $\forall n \geq a(n \in S \to n + 1 \in S)$, I wrote the following:

Suppose n belongs to S for all $n \geq a$,

... ...

then n+1 also belongs to S.

I'm afraid the beginning of the proof sounds like a complete statement stating that $\forall n \geq a(n \in S)$, which is clearly not what I intended.

I have thought about some alternatives: the book states that wherever some integer $n \geq a$ belongs to S, then n+1 also belongs to S. However, I am unsure about the usage of "some" as it often implies existence. Alternatively, I was thinking about writing suppose n belongs to S where $n \geq a$, then we want to show n+1 is also in S, and the proof continues. However, I felt that this doesn't clearly identify that the statement holds for all n. Additionally, I'm less inclined to write the phrase "we want to show".

So what word choice should I adopt to write this proof? In general, how should I translate the above formula written with logic symbols to standard English?

Answer 1

$\forall x\in D~(x\in A\to x\in B)$ says "For every thing in D, if that thing in A , then that thing is in B."

More briefly, "Everything in D that is in A will be in B"

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Take any thing in D. Suppose it is in A. ...(insert a valid derivation).... So such a thing will be in B.

Answer 2

$\forall n{\geq}a\;(n \in S \to n + 1 \in S)\tag1$

• Suppose $n$ belongs to $S$ where $n \geq a$, then we want to show $n+1$ is also in $S.$

This is pretty okay (though I'd replace “where” with “and”), as it suggests the equivalent statement $$\forall n\;(n \in S\;\text{and}\;n\geq a\;\implies\;n + 1 \in S).\tag2$$ In fact, noting that all three statements $$\forall n{\in}S\;(n\geq a\;\implies\;n + 1 \in S)\tag3$$ are equivalent to one another, here are some alternatives (and more at ‘Any’ versus ‘arbitrary’):

• Let $n$ be arbitrary and suppose that $n\in S$ and that $n\ge a\ldots$
• Let $n$ be an arbitrary element of $S,$ and suppose that $n\ge a\ldots$
• Let $n$ be an arbitrary element of $S$ such that $n\ge a\ldots$

Also: How to interpret “let... suppose” in mathematics?

However, I feel that this doesn't clearly identify that the statement holds for all $n.$

Yes, because at this point you are referring to a representative value of $n.$ Then, when wrapping up the proof, you conclude generally that “for each $n\ldots$”, by relying on Universal Introduction.

Additionally, I'm less inclined to write the phrase “we want to show”.

I think the phrase is reader-friendly!

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Addendum

@ryan I think the main part I was missing is the “arbitrary element”.

The $n$ in your suggestion is tacitly understood as arbitrary.

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user21820:

Contrary to ryang's post, I recommend avoiding "let" for ∀-quantification.

This really is a non-objection: it is the word “arbitrary” (tacit or otherwise)—not the word “let”— that signals ∀.

Answer 3

Your last idea is already very close.

Suppose $n\geq a$ is such that $n\in S$. Then ⟨your proof here⟩, and thus $n+1\in S$.

Varying $n$ in the above argument, we see that, for any $n\geq a$, the implication ${n\in S}\to{n+1\in S}$ holds.

Answer 4

Contrary to ryang's post, I recommend avoiding "let" for $∀$-quantification. If you want actual idiomatic English, you should not use "let". The idiomatic alternative is:

Consider/take any n ≥ a that belongs to S. Then ... Thus n+1 also belongs to S.
Therefore by induction every n ≥ a belongs to S.

The evidence is clear; you can see people telling you how to interpret "let" in mathematics, showing that the common manner it is used in mathematics is not idiomatic English.