Assignment IV (Temporal Logic)


Due Nov 26, 1997

Please no teamwork for this assignment!

Consider the program (forever C) where:

C =    x = (load = 1 -> In1 | x)
       y = (load = 1 -> y | not x)
       done = (load = 1 -> 0 | 1)
Prove the following temporal statement using the axioms and inference rules in the book. You may use other axioms and rules if you first show their soundness:
{ forever C }
IMPLIES Box { done=0 IMPLIES load=0 }
IMPLIES Box ( {load=1 AND In1=a } 
              IMPLIES DIAMOND { x=a AND y=not a AND done=1 })
Extra: Can you prove a stronger statement?
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