and

user=> (and true true)
true

user=> (and true false)
false

user=> (and false false)
false

user=> (and '() '())
()

user=> (and '[] '[])
[]

user=> (and 0 1)  ; Note that this is *not* bitwise 'and'
1

user=> (and 1 0)
0

;; See examples for "if" explaining Clojure's idea of logical true
;; and logical false.

; Note that, and does not evaluate if the first value is false
user=> (and false nil)
false

user=> (and nil false)
nil

user=> (and false (println "foo"))
false

user=> (and (println "foo") false)
foo
nil

user=> (and nil nil)
nil

; From the Clojure 1.9 source code.
(defn qualified-keyword?
  "Return true if x is a keyword with a namespace"
  [x] (and (keyword? x) (namespace x) true))

; Note how the return value of and is value of the last expression.
user=> (qualified-keyword? :hi/there)
true

; If we instead define the function as:
(defn qualified-keyword?
  [x] (and (keyword? x) (namespace x)))
; we get the namespace as return value:
user=> (qualified-keyword? :hi/there)
"hi"