Match each schema with the corresponding rule from predicate logic to prove it, or
indicate why it is not a theorem
p ├ (∀x)(p v x = y)
((∀x)φxz) ^ x = y ├ (∀y)x = y
(∀x)(hgx = y ^ T) ├ hgx = y ^ T
*Note there are more options to choose from than needed, and you can use any
option as many times as you wish (or not at all)
None
strong generalization
weak generalization
specialization