Notation: I use - \x denotes lambda abstraction (1) Prove that, for a closed lambda-term M, if M has a head-normal-form, then there is a sequence of terms P1, ..., Pn such that M P1 ... Pn = I (This is where the terminology "solvable" comes from.) (2) Prove directly (that is: without using B"ohm's Theorem) that the equation c1 = c2 is inconsistent. (c1 is Church numeral "one": \f.\x.f x and c2 is Church numeral "two": \f.\x.f(f x).) (3) Define A := \x. x y (x x) and M := A A Draw the B"ohm tree of M. (4) Suppose that the term B satisfies B = x B B. Draw the B"ohm tree of B. (5) (a) Give a term P that has the following B"ohm tree: x | x |\ x y | x |\ x y | etcetera (b) (Hard) Give a term Q that has the following B"ohm tree: x | x |\ x y | x |\ x y | \ x y |\ x y | \ x y |\ \ x y y | \ etcetera