2 deleted 47 characters in body edited Jan 28 at 16:04 Draconis 6,32711 gold badge1010 silver badges2121 bronze badges The answer here is the same as in the other question: one thing is missing here! Your addition result should be: $$3 + 4 = \lambda g . \lambda z . 3 g (4 g z) = \lambda g . \lambda z . 7 g z$$ Note that $$g$$ is now a lambda parameter, not a free variable! And as such, it doesn't get alpha-transformed. So now if you want to apply this to something, it'll get substituted in the same everywhere: $$7 q r = (\lambda g . \lambda z . 7 g z) q r = q q q q q q q r$$ The answer here is the same as in the other question: one thing is missing here! Your addition result should be: $$3 + 4 = \lambda g . \lambda z . 3 g (4 g z) = \lambda g . \lambda z . 7 g z$$ Note that $$g$$ is now a lambda parameter, not a free variable! And as such, it doesn't get alpha-transformed. So now if you want to apply this to something, it'll get substituted in the same everywhere: $$7 q r = (\lambda g . \lambda z . 7 g z) q r = q q q q q q q r$$ The answer here is the same as in the other question: one thing is missing here! Your addition result should be: $$3 + 4 = \lambda g . \lambda z . 3 g (4 g z) = \lambda g . \lambda z . 7 g z$$ Note that $$g$$ is now a lambda parameter, not a free variable! So now if you want to apply this to something, it'll get substituted in the same everywhere: $$7 q r = (\lambda g . \lambda z . 7 g z) q r = q q q q q q q r$$ 1 answered Jan 28 at 3:56 Draconis 6,32711 gold badge1010 silver badges2121 bronze badges The answer here is the same as in the other question: one thing is missing here! Your addition result should be: $$3 + 4 = \lambda g . \lambda z . 3 g (4 g z) = \lambda g . \lambda z . 7 g z$$ Note that $$g$$ is now a lambda parameter, not a free variable! And as such, it doesn't get alpha-transformed. So now if you want to apply this to something, it'll get substituted in the same everywhere: $$7 q r = (\lambda g . \lambda z . 7 g z) q r = q q q q q q q r$$