Etydhv

I am following this video tutorial, starting 6th minute

I would like to parse the following sentence

square blue or round yellow pillow. 

For now I am interested in only how square and blue are combined.

In particular with start with the following representation

square -> ADJ: lambda x. square(x)
blue -> ADJ: lambda x. blue(x)

Next step is we raise types:

square -> N/N: lambda f. lambda x. f(x) / square (x)
blue -> N/N: lambda f. lambda x. f(x) / blue (x)

Now we create representation for square blue. I indicate substitution by brackets

lambda x. [ lambda f. lambda x. f(x) / blue (x) ] (x) / square (x)

Next I simply substitute z for x everywhere outside of square brackets, so that we do not confuse different xs.

lambda z. [ lambda f. lambda x. f(x) / blue (x) ] (z) / square (z)

Next I push z into square brackets

 lambda z. lambda x. z(x) / blue (x) / square (z)

This is different from what stated in the lecture:

lambda z. lambda x. z(x) / square (x) / blue (x) 

Where did I make a mistake?

I'll leave the lambdas to you, but you might like to know the syntactic structure. It is a RNR (right node-raising) construction, with "pillow" the raised node. The intonation I think makes that obvious. Thus:

[square blue GAP or round yellow GAP] (pillow), where the noun "pillow" fills the GAP.

"square", "blue", "round", "yellow" are adjectives which modify nouns or modified nouns.

If I am not mistaken, your computation is correct, and the video simply shows the wrong result.

The order square(x) ^ blue(x) comes from the fact that squared is applied to blue where blue will be substituted for f in the term f(x), which comes before square(x) in the conjunction.

To change the order in which the terms will appear in the conjunction without changing the order in which the terms are applied to each other (of course you could also do the backward composition blue square so square goes in for f in blue, but that wouldn't be in line with the intended forward composition), one could simply change the order of the expressions in the definition of square to

square -> N/N: lambda f. lambda x. square (x) ^ f(x)

which changes the rule for type raising to

lambda x. g(x) -> lambda f. lambda x. g(x) ^ f(x)

(λf.[λx.[f(x) ^ square(x)]])(λf.[λy.[f(y) ^ blue(y)]]) reduces to
(λx.[(λf.[λy.[f(y) ^ blue(y)]])(x) ^ square(x)]) which reduces to
(λx.[(λy.[x(y) ^ blue(y)]) ^ square(x)]),

not (λf.[λx.[f(x) ^ square (x) ^ blue (x)]]).

I have no idea how they arrive at this result, it must be a mistake.