Answer:
Step-by-step explanation:
Reduction to normal from using lambda-reduction:
The given lambda - calculus terms is, (λf. λx. f (f x)) (λy. Y * 3) 2
For the term, (λy. Y * 3) 2, we can substitute the value to the function.
Therefore, applying beta- reduction on "(λy. Y * 3) 2" will return 2*3= 6
So the term becomes,(λf. λx. f (f x)) 6
The first term, (λf. λx. f (f x)) takes a function and an argument, and substitute the argument in the function.
Here it is given that it is possible to substitute the resulting multiplication in the result.
Therefore by applying next level beta - reduction, the term becomes f(f(f(6)) (f x)) which is in normal form.
3 should be added to the tiles
Answer:
3
Step-by-step explanation:
Break the problem up into two
Ignore the -1 for now
2+2=4 because if you separate 2 and 2, you get four ones, therefore it's four
Now that you have four, subtract 1
And you got 3
Hmm I would say x=q
Hope this helps u <3
