Slight edits from lcnr's review

Author: jackh726

src/divergence.md

@@ -23,6 +23,9 @@ The following language constructs unconditonally produce a _diverging expression
 r[divergence.diverging-expressions.conditional]
 In a control flow expression, if all arms diverge, then the entire expression also diverges.
 
+r[divergence.diverging-expressions.place-expressions]
+A place expression of the type [`!`](./types/never.md) is considering _diverging_ only if it is read from.
+
 r[divergence.fallback]
 ## Fallback
 If a type to be inferred is only unified with diverging expressions, then that type will be inferred to be `!`.

src/expressions/block-expr.md

@@ -61,7 +61,7 @@ assert_eq!(5, five);
 ```
 
 r[expr.block.type.diverging]
-However, if there are any values unconditionally created within a block that are [diverging](../divergence.md), then the block itself is considered diverging.
+A block is itself considered to be [diverging](../divergence.md) if all reachable control flow paths contain a [diverging expression](../divergence.md#r-divergence.diverging-expressions).
 
 r[expr.block.value]
 Blocks are always [value expressions] and evaluate the last operand in value expression context.

src/expressions/match-expr.md

@@ -97,7 +97,7 @@ r[expr.match.binding-restriction]
 Every binding of the same name must have the same type, and have the same binding mode.
 
 r[expr.match.type]
-The type of the overall `match` expression is the least upper bound of the individual match arms.
+The type of the overall `match` expression is the [least upper bound](../type-coercions.md#r-coerce.least-upper-bound) of the individual match arms.
 
 r[expr.match.type.diverging.empty]
 If there are no match arms, then the `match` expression is diverging and the type is [`!`](../types/never.md).