E. Lemma, 5.1. For any term t and base term b in ? CA , ?(t

. Proof, We proceed by structural induction over t ? ? CA ? t = x. Then ?(x[b/x]) = ?(b) = x[?(b)

@. and =. ?y, (?y : U.t ? [b/x]) and this is equal to ?y : U.?(t ? [b/x]) which by the induction hypothesis is equal to ?y, (t ? ))[?(b)/x] = ?(?y : U.t ? )[?(b)/x]

@. and =. ?x, r[b/x]) = ?X.?(r[b/x]) which by the induction hypothesis is equal to ?X.?(r)[?(b)

@. and =. R@u, Then ?(r@U = ?(r[b/x]@U ) = ?(r[b/x])@U which by the induction hypothesis is equal to ?(r)[?(b), @U = ?(r)@U [?(b)/x] = ?(r@U )[?(b)/x]

@. and =. R+u, = ?(r[b/x])+?(u[b/x]) which by the induction hypothesis is equal to ?(r)[?(b)

E. Lemma, For any terms t 1 , t 2 in ? add

. Proof, Notice that the only case where these terms are different is when one of the addends reduces to 0, in such case, their normal forms coincides, making it possible to compare in this way

. Proof, We proceed by structural induction over t ?? CA

=. ?x, By the induction hypothesis ?(r)? A ?(r?), so ?(?X.r)? A = ?X.?(r)? A ?X.?(r?) = ?(?X.r?)

. Proof, Rule by rule analysis. Elementary rules ? Rule 0

@. Application and . Rule, ? (t) u + (r) u. (t + r) u = (t + r) u ? v (t) u + (r)

@. Rule, Analogous to previous case Beta reductions ? Rule (?x : U.t) b ? t[b/x]. (?x : U.t) b = (?x.t) b. Since base vectors are translated into base vectors

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