Deriving expression type in SML

Question

I'm kinda new to sml and i'm trying to understand how to derive the expression type.
I read Deriving type expression in ML and I'm trying to follow what he did to derive my expression, but I can't understand one passage.

The expression I'm trying to derive is:

fn x => fn y => fn z => z(z(y x))

What I'm doing is as follow:

fn x : T1 => fn y : T2 => fn z : T3 => z(z(y x)) : T4

At this point it should be

T2 = T1 -> T5                       //y x
T3 = T5 -> T6                       //z(y x)
T3 = T6 -> T7                       //z(z(y x))

which is where I think I'm doing the mistake.

Also there should be a constraint like

T7 = T4 or T6 = T4.

I'm not sure nor clear on this point tho.

Thanks in advance for the help.

score 1 · Accepted Answer · answered Apr 03 '14 at 21:24

1

You have already assigned T4 to z(z(y x)), so we can use it in the last equation.

T3 = T5 -> T6                       //z(y x)
T3 = T6 -> T4                       //z(z(y x))

Since T5 -> T6 = T6 -> T4, it follows that T5 = T6 and T6 = T4.

answered Apr 03 '14 at 21:24

n. m. could be an AI

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Thanks for the answer, could you explain how T5 = T6 and T6 = T4? Is not very clear to me. Sorry I'm not sure if i'm dumb or something. – AscaL Apr 03 '14 at 21:39
One compound thing `T5 -> T6` equals another compound thing `T6 -> T4`, their corresponding parts are equal too. – n. m. could be an AI Apr 03 '14 at 23:06
Thanks a lot, I had a doubt about this but I solved it. Thank you. – AscaL Apr 04 '14 at 00:20

Deriving expression type in SML

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