Simplifying function output not exactly as expected

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I am trying to write a function that will simplify an arbitrary list of boolean expressions, but my function fails certain tests.

  (defn sim
  [expression]
  (if (some true? expression)
    true
    (if (= (count expression) 2)
      (if (some false? expression)
        false
        (if (map simple-symbol? expression)
          (if (= (count expression) 2)
            (drop 1 expression)
            expression)))
      (if (some simple-symbol? (drop 1 expression))
        (filter simple-symbol? expression)))))

When I call using (sim '(or x false)) I expect the output to be (x), but instead it returns (or x). Conversely, when I call using (sim '(or x)) my output is (x) as expected.

How about something along these lines? This is only for or, but I'm sure we can do the same for and and other boolean operators.

(defn simplify-or [exp]
  (let [op (first exp)
        args (rest exp)]
    (when (= op 'or)
      (let [nf-args (filter #(symbol? %) args)]
        (if (some true? args)
          true
          (case (count nf-args)
            0 false
            1 (first nf-args)
            (concat (list op) nf-args)))))))

Results:

(simplify-or '(or false))       
=> false

(simplify-or '(or true))       
=> true

(simplify-or '(or x true y false))       
=> true

(simplify-or '(or x y false))       
=> (or x y)

my output is not simplified to one number, results in the expected output. with the parameters for the variables. t = -​120.07208, m = -0.80227, a = 0.77755, b = 0.00161, c = 1.46526E-6, d = 7.15449​E-8,  This is the expected behavior of the SIMPLIFY function in the Symbolic Math Toolbox. By default, the function tries to make as few assumptions about the symbolic variables as possible. Considering a simple example, the expression 'log(exp(x))' cannot always be simplified to the expression 'x'.

(defn simplify-or
  [[op & args]]
  (let [args-without-false (remove false? args)]
    (cond
      (some true? args-without-false) true
      (= 1 (count args-without-false)) args-without-false
      (empty? args-without-false) false
      :otherwise (conj args-without-false op))))

(simplify-or '(or x false y))
#=> (or x y)
(simplify-or '(or x))
#=> (x)
(simplify-or '(or x true y false))
#=> true
(simplify-or '(or false false)
#=> false

My concern is some inconsistency here, what is (x)? why not just x? The same way as we return just true or false.

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(require '[clojure.walk :as w])

(defmulti dispatch first)

(defmethod dispatch 'or
    [[op & args]]
    (if (contains? (into #{} args) true)
        true
        (case (count (remove false? args))
            0 false
            1 (first (remove false? args))
            (cons op (remove false? args)))))

(defmethod dispatch 'and
    [[op & args]]
    (if (contains? (into #{} args) false)
        false
        (case (count (remove true? args))
            0 false
            1 (first (remove true? args))
            (cons op (remove true? args)))))

(defmethod dispatch :default [x] x)

(defn simplify [x]
    (prn (w/postwalk (fn [x]
                         (if (and (list? x) (seq x))
                             (dispatch x)
                             x))
                     x)))

(simplify '(or x false))
(simplify '(or x (or y false) z false))
(simplify '(or x (or y false) (and z false)))
(simplify '(or x false y))
(simplify '(or x))
(simplify '(or x (or x true y false)))
(simplify '(or false false (and true true)))

Why is the simplify function not simplifying?, This is the expected behavior of the SIMPLIFY function in the Symbolic Math Toolbox. By default, the function tries to make as few assumptions about the  Since Sqrt is a function, it must be defined with a branch cut (this is not a shortcoming of Mathematica, just a result of requiring functions to be single-valued). The cut used by Mathematica means that phases of complex arguments are moved into the interval before applying the rule .

Simplifying Boolean Expressions and Conditionals, It is alway possible to replace the pattern if (E) {return true;} else {return false;} by the any boolean expression E. We expect you to make this simplification in your code. Just as it is possible to simplify algebraic expressions by using rules like valid simplifications; for this reason, they are not included in the above table:  However, it does not simplify the statement. Indeed, the result requires two evaluations of boolExp1 where the initial expression only required one! However, if at least one of boolExp2 or boolExp3 is true or false , then the boolean expression simplifications from above can lead to a simpler result.

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