Update to 'given' syntax for inferable parameter

docs/docs/reference/contextual/derivation.md

@@ -40,7 +40,7 @@ The generated typeclass instances are placed in the companion objects `Labelled`
 
 A trait or class can appear in a `derives` clause if its companion object defines a method named `derived`. The type and implementation of a `derived` method are arbitrary, but typically it has a definition like this:
 ```scala
-  def derived[T] with Generic[T] = ...
+  def derived[T] given Generic[T] = ...
 ```
 That is, the `derived` method takes an inferable parameter of type `Generic` that determines the _shape_ of the deriving type `T` and it computes the typeclass implementation according to that shape. An implied `Generic` instance is generated automatically for any type that derives a typeclass with a `derived`
 method that refers to `Generic`. One can also derive `Generic` alone, which means a `Generic` instance is generated without any other type class instances. E.g.:
@@ -218,7 +218,7 @@ trait Eq[T] {
 We need to implement a method `Eq.derived` that produces an instance of `Eq[T]` provided
 there exists evidence of type `Generic[T]`. Here's a possible solution:
 ```scala
-  inline def derived[T] with (ev: Generic[T]): Eq[T] = new Eq[T] {
+  inline def derived[T] given (ev: Generic[T]): Eq[T] = new Eq[T] {
     def eql(x: T, y: T): Boolean = {
       val mx = ev.reflect(x)                    // (1)
       val my = ev.reflect(y)                    // (2)
@@ -315,7 +315,7 @@ calling the `error` method defined in `scala.compiletime`.
 **Example:** Here is a slightly polished and compacted version of the code that's generated by inline expansion for the derived `Eq` instance of class `Tree`.
 
 ```scala
-implied [T] with (elemEq: Eq[T]) for Eq[Tree[T]] {
+implied [T] given (elemEq: Eq[T]) for Eq[Tree[T]] {
   def eql(x: Tree[T], y: Tree[T]): Boolean = {
     val ev = the[Generic[Tree[T]]]
     val mx = ev.reflect(x)