At first glance this looks like an easy problem. But actually building the different parse trees for what I would expect here, and in all cases it should be True.
So that is to say:
a == (b is c) --> ==
/ \
a is
/ \
b c
(a == b) is c --> is
/ \
== c
/ \
a b
a == b is c --> ==_is
/ | \
a b c
Another hypothesis is that == and is are not pure functions that operate on values, but are special operators that operate on syntax, and treat a parenthesized expression differently from an unparenthesized one.
(In what way does this sort of thing belong in a self-proclaimed newbie-friendly language? This has to be a bug.)
If you're working in TXR Lisp, you can use the meq, meql and mequal functions for three or more argument equality.
The call:
(meq a b c ...)
(and (eq a b) (eq b c) ...)
(or (eq a b) (eq a c) (eq a d) ...)
These functions can be called with only one argument, in which case they yield false, just like (or).
These functions are very useful in cond statements that include some tests that prevent conversion into caseq/caseql/casequal.
(cond
((meql x 1 2 3) ... ) ;; x is one of 1 2 3
((> x 10) ...) ;; x is greater than 10
(t ...)) ;; otherwise