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Problem 1.5
Friday, August 23, 2019; ago; Download .md

Using applicative order, the code for (test 0 (p)) never terminates because it infinitely expands into itself.

(test 0 (p)) 

(test 0 (p)) 

(test 0 (p)) 

However, for normal eval it evaluates step-by-step to 0.

(test 0 (p)) 

(if (= 0 0) 0 (p)) 

(if #t 0 (p)) 

0