Aha, thanks!
A good mnemonic is that that if X and Y are finite types[0], sum, product and exponent types are just doing the relevant arithmetic on the types.
(Bool->Bool,Uint8+()) is just a verbose way of saying (2^2 * (256+1)), where 1,2, and 256 are types with the appropriate number of elements.
The cases involving infinite types are just a (straightforward?) generalization of that.
0: eg 0=noreturn [1],1=void,2=Bool,256=uint8_t
1: Much like in set theory, zero is the empty set.
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A good mnemonic is that that if X and Y are finite types[0], sum, product and exponent types are just doing the relevant arithmetic on the types.
(Bool->Bool,Uint8+()) is just a verbose way of saying (2^2 * (256+1)), where 1,2, and 256 are types with the appropriate number of elements.
The cases involving infinite types are just a (straightforward?) generalization of that.
0: eg 0=noreturn [1],1=void,2=Bool,256=uint8_t
1: Much like in set theory, zero is the empty set.