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25 Apr 2002: Next, we perhaps note that if b=a. 2. then we have the equation x. ... a-3t)y. 2. (b-2at3t. 2. y) c-btat. 2. t. 3. This gives us a'=a-3t and b'=b-2at3t.

27 Aug 2002: 2H0&<(?:/B?&, B>,B+:DY&H]r%3&$0XB?3&7+32H(&$0(:/B?&<:/(¤Ç 4r&<(?:/DA3., B+&,X03&$4ã98, 0152HDIH, 9?:, ;.&YL N KÇJOF4T:/B?IH,. ... 015t2H9TÚ wy (?&$B Ú9 f Æ ÚûK£ÇYO u Ç w »# È }%'&C7$, 0&Z=.2:/B)B?&<£,7"B)B+-, B>B+3&72H0-5:/B+:20-,.

26 Sep 2002: Next, use the other &E axiom:. [[A & B]] [[B]] [[B]] [[C]][[A & B]] [[C]]. ... A. [[(A ) ]] [[A]]) (. B. [[B (B )]]). Here the premise is a schematic rule.

11 Nov 2002: k (a, b) | b, which contradicts (a, b) being the highest common factor. ... Proof: () (a, b) | a and (a, b) | b, so (a, b) | (ax by) = c.

15 Mar 2002: a) Show that, if A k A′ and B k B′, then A B k A′ B′.(b) Show that, for n,m k, nA k mA. ... Suppose that for each i < n, (A,ai,ai1) 3p(B,bi,bi1). Show that (A, a) p (B, b) (i.e.

26 Sep 2002: Γ a AΓ,b B a A. The product introduction rule is handled like -introduction. ... Γ A type Γ, priΓ,b,B A b(priΓ,b,B) B(priΓ,b,B)Γ lambda(b) Π(A,B). In Isabelle, the Type Theory parser, read rule, is

18 Sep 2002: U4Z+[0:J81:<;1G/ '9Ja' m /HJKml'54p)L<;1G/ '9Jq 12GgK1 D 02'9Y54lW[=WX'(e5+"7(e"'(WY5)I+J8130;79e5<81:<Vb#b#b#bq /HJ8021T4rYWX1G/ '9JK0 D ... 13'Ô+JWX'(e5+"b% "b c@b A &(xs YK0=PpªV> /HJ[0GÄg ª O9?

26 Sep 2002: A B. ABa : A b : B〈a,b〉 : AB. 14 Chapter 2. ... n : Nat b : B(zero). [x : Nat ; y : B(x).

26 Sep 2002: A,B | B A,B | AA,B | B A :right. AB | B A :left. ... R(?a,f(?a)) and R(?b,? b). 1.4 Parameters in quantifier rules.

26 Sep 2002: α = [. , 0, 0, a, 0, b, 1 a b, 1, 1 a b, b, 0, a, 0, 0,. ] ... α(1) = 3[. , 0, 0, b 13,13 b, b, 1 2b, b, 1.