We have an initial set of premises $S$.

We are given or observe or assume sentence(s) $A$ is/are true.

We can then prove $B$.

Formally, $S \cup \left\{A\right\} \vdash B$.

Shouldn't it be an intersection? Only when we have both $S$ and ${A}$ can we prove $B$. Not when we have $S$ or ${A}$, which is what my mental picture of what Union is.

  • $\begingroup$ You are "using" $\lor$ in the wrong way ... The set $S \cup \{ A \}$ is not a set that is "equal" to $S$ or to $A$ but is the set of those objects $x$ that belong to $S$ or belong to $\{ A \}$ : $x \in S \cup \{ A \}$ iff $x \in S \lor x \in \{ A \}$ i.e. $x \in S \cup \{ A \}$ iff $x \in S \lor x = A$. $\endgroup$ – Mauro ALLEGRANZA Jun 28 '15 at 19:16

The English word "and" has two different senses, with different translations into logic.

When you say you ate soup and salad, you don't mean that you ate something that is both soup and salad. You mean you ate a soup, and you also ate a salad. Your lunch, then, consisted of the union $$\{\text{Soup}\} \cup \{\text{Salad}\},$$ not the intersection.

That is the sense that is being used here. When you have a set with both $S$ and $A$, you have $$S\cup \{A\}$$ which contains $S$ and also includes $A$.

Or consider "I visited France and Japan this summer." You don't mean you visited the intersection of France and Japan; that intersection is empty.

Also note that the intersection is clearly wrong here, since $S\cap \{A\}$ is equal to either $\emptyset$ or $\{A\}$, so doesn't contain anything from $S$ other than $A$.

  • $\begingroup$ I never realized "and" is ambiguous until now... $\endgroup$ – user541686 Jun 28 '15 at 11:47

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