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The syntax of the language is centered around M-expressions, which is equivalent to LISP's s-expression.

While the evaluator behaves differently than many LISPs , mainly being term rewrite with pattern matching, with a global rule base, I believe both are Turing complete, and furthermore close enough to easily be extended so each can mutate to the other. A big swath of Scheme follows the substitution model anyway so they are very close indeed.

But are they close enough ? Can Mathematica be considered a LISP , albeit with its own peculiar evaluator.

Another way of stating this question is : What is the essence of a LISP ? what makes a language based of an abstract syntax tree with an evaluator a LISP or not ?

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    $\begingroup$ Have you seen Why did the Mathematica Language choose term rewriting instead of the Lambda Calculus as its basis? $\endgroup$
    – MarcoB
    Commented Nov 4, 2023 at 0:31
  • $\begingroup$ @MarcoB Yes, and it did not satisfy me, because you CAN do lambda calculus in Mathematica. $\endgroup$
    – Ehab Shoubaki
    Commented Nov 4, 2023 at 2:42
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    $\begingroup$ I think they feel close enough. One other difference, however, is that M- is an infinite loop evaluator, while LISPs are mostly single step evaluators. It's probably simpler to make a single step evaluator behave like an infinite loop evaluator then the other way around. $\endgroup$
    – user21
    Commented Nov 4, 2023 at 5:47

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