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Nov 9, 2017 at 9:07 comment added Peter Taylor Is this higher order unification? The question seems to be about the untyped lambda calculus rather than the simply typed lambda calculus.
Nov 9, 2017 at 7:32 comment added BarbaraKwarc Thanks for the link to the paper too, I'll check it out, and I'll accept your answer in a couple of days to give other people a chance too.
Nov 9, 2017 at 7:31 comment added BarbaraKwarc Yup, seems like it :) The funny thing is, I almost got that solution, but from some reason I thought that calling $(\lambda x.M)$ on something would "execute it", messing up the expression :q What I missed is that we can substitute the variable with another variable bound outside.
Nov 9, 2017 at 6:14 history answered Joey Eremondi CC BY-SA 3.0