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26 Sep 2002: i,j) evnodd A b = ((i,j) A (ij) mod 2 = b)evnodd (A B) b = evnodd A b evnodd B b. ... goalw thy [evnodd_def] "evnodd (A - B) b = evnodd A b - evnodd B b";by (Blast_tac 1);qed "evnodd_Diff";.

26 Sep 2002: parts (sees Spy evs) B=B’ P=P’. Although it is not used in later proofs, unicity of nonces is important. ... Hash {|Key(shrK A), Agent A, B, NA, P|}. parts (sees Spy evs) B=B’ P=P’.

13 Sep 2002: Step 1. Eliminate and by repeatedly applying the following equivalences:. A B ' ( A B) (B A). ... A A A. B A B. A B B A. (B C) ( A B A C).

11 Nov 2002: Then either B B.x B or C C.x C. In both cases B B. ... Then B B. C C.x B C so B B.x B ) (c CC).

17 Dec 2002: 3 = # / " -# <= )B+ # 2" 84-"$/, % ;$/ <& 2" =! =! 4( "4 1B 8. B "" " B+ "0%'(" , 8 7 ,-G%@ 3 , 78 ,"G" , 8: ,BGA"9. , ... J /L J ,"8 ,L. ; B ( ) 7 >- B+ # & B # #! $/ (O ". %"<; 7'" B )% " 47 >- ,08 DB!

24 Sep 2002: j j, p s b b w j p q y3b 2 q10 b b y3b 2 w 2 j p 2Ž. ... G s b 2 q b 2 r q b b r ,Ž.

26 Sep 2002: a =α a. a =α b. b =α a. a =α b P (b). ... arithmetic is developed up to the theorem a mod b (a/b) b = a.Reflection works well in higher-order reasoning.

31 Oct 2002: n2. (b) Let X be any random variable which takes on values 0,1,2,. ... a) Determine the equilibrium points. (b) Determine the stability of each of the equilibrium points.

26 Sep 2002: variable; the printer chooses a unique name. A loose bound variable is printed B.n. ... op thenp: ’a parser ’b parser -> (’a’b)parser. Composition of functionals: parses one phrase, then another.

28 Mar 2002: X <Y6N12Z[2JOL2> &2 Q]2_ ]a bcO$$d2JOL2> &2 Qef<@;)A)B)B2B)AN C bghd&.ji# kl5& m@onqp$,7&h.$M srt< ; A)B)B)BA1CR ] ; ... $M. { rt t;A)B)B)B2AN OR ];2AB)B)B)AN]STM= <p$#,&. $KJ#),)h1&. $O 9 }S@' ] bcO$$#)dJ#),)h1&.