STEPN:

    ---------------- step0
    H; s -0-> H; s



    H; s -N-> H'; s'   H'; s' --> H''; s''
    ------------------------------------------ stepS
    H; s -N+1-> H''; s''


Divergence:

Theorem: while 1 skip always deverges, i.e.,

Restated: For all H and n, there exists H' and s' such that
    H; while 1 skip -->n H'; s'

TRY #1
Proof: By induction on n, the number of steps

Base case: 0 steps
   After 0 steps, H'=H and s' = while 1 skip. Done.

Inductive case n > 0
    By induction, there exists an H'' and s'' such that
    H; while 1 skip -->n-1 H'';s''

    So it suffices to show: for all H'';s'' there exists and H' and s'
    such that H'';s'' --> H';s'

    OOPS almost works except if s'' is skip
    (But it's kind of ridiculous because needed that every program
     can always take some step.)

TRY #2
Stronger theorem: For all H and n, there exists H' and s' such that
   H; while 1 skip -->n H'; s' and s' is not skip

Base case: 0 steps
   Like before, let H'=H and s'=while 1 skip and notice while 1 skip
   is not skip

Inductive case: n > 0
    By induction, there exists an H'' and s'' such that
    H; while 1 skip -->n-1 H'';s'' and s'' is not skip

    So it suffices to show: for all H'';s'' where s'' is not skip,
    there exists and H' and s' such that H'';s'' --> H';s'

    cases for all kinds of s and I'm done, right?

    OOPS NO, also have to show s' is not skip!
      skip;skip
      x:=e
      if 7 skip s
    once again, this induction hypothesis is too weak!

TRY #3

Stronger theorem: For all H and n,  
    H;while 1 skip -->n H; while 1 skip

Base case: 0 steps
    Yes, after 0 steps we have H and while 1 skip

Induction: n > 0
     By induction H;while 1 skip -->n-1 H; while 1 skip
     So we just need that H;while 1 skip --> H; while 1 skip

     OOPS That's not true: it steps to H; if 1 (skip; while 1 skip) skip

     In fact, the theorem is not true.

     Lesson: If you strengthen too much you try to prove something 
     false and that better not work.

TRY #4 (correct)

Stronger theorem: For all H and n, 
    H;while 1 skip -->n H; s
   where s is one of:
      while 1 skip  (call this s1)
      if 1 (skip; while 1 skip) skip (call this s2)
      skip ; while 1 skip (call this s3)

By induction on n:
 
Base case n=0:
   Let s be s1

Induction case n > 0:
   By induction H;while 1 skip -->n-1 H;s where s is s1, s2, or s3.
   Proceed by cases:
     case: If s is s1, then H;s1 --> H;s2.
     case: If s is s2, then H;s2 --> H;s3.
     case: If s is s3, then H;s3 --> H;s1.