Document edge case eqv behaviour…

…when comparing lazy iterables

Author: zoffixznet

doc/Language/operators.pod6

@@ -1529,6 +1529,16 @@ the same, i.e. from the same type and (recursively) contain the same values.
     say 1 eqv 2;                    # OUTPUT: «False␤»
     say 1 eqv 1.0;                  # OUTPUT: «False␤»
 
+Note that lazy L«C<Iterables>|/type/Iterable» cannot be compared, as
+they're assumed to be infinite. However, the operator will do its best and
+return C<False> if the two lazy C<Iterables> are of different types or
+if only one C<Iterable> is lazy.
+
+    say (1…∞) eqv (1…∞).List; # Both lazy, but different types;   OUTPUT: «False␤»
+    say (1…∞) eqv (1…3);      # Same types, but only one is lazy; OUTPUT: «False␤»
+    (try say (1…∞) eqv (1…∞)) # Both lazy and of the same type. Cannot compare; throws.
+        orelse say $!.^name;  # OUTPUT: «X::Cannot::Lazy␤»
+
 The default C<eqv> operator even works with arbitrary objects. E.g., C<eqv>
 will consider two instances of the same object as being structurally
 equivalent: