Proof: Byte 3 Stays the Same 18
Let's prove the following theorem:
if the following are true:
    
    - instruction #6 is 
addi dst=2 src=3 imm=0 - the PC at time 18 = 6
 - value of cell 3 at time 18 = 3
 
then value of cell 3 at time 19 = 3
Instructions
        
    | Memory Cells | 
|---|
| Program Counter | Time | 
|---|---|
| 0 | 0 | 
  
  
LW Computer Simulator
Proof:
  
      
      Given
      
    
    
      
  
  
| 1 | instruction #6 is addi dst=2 src=3 imm=0 | 
      
|---|---|
| 2 | the PC at time 18 = 6 | 
| 3 | value of cell 3 at time 18 = 3 | 
| # | Claim | Reason | 
|---|---|---|
| 1 | not (3 = 2) | not (3 = 2) | 
| 2 | value of cell 3 at time (18 + 1) = value of cell 3 at time 18 | if instruction #6 is addi dst=2 src=3 imm=0 and the PC at time 18 = 6 and not (3 = 2), then value of cell 3 at time (18 + 1) = value of cell 3 at time 18  | 
  
| 3 | 18 + 1 = 19 | 18 + 1 = 19 | 
| 4 | value of cell 3 at time (18 + 1) = value of cell 3 at time 19 | if 18 + 1 = 19, then value of cell 3 at time (18 + 1) = value of cell 3 at time 19 | 
| 5 | value of cell 3 at time 19 = value of cell 3 at time 18 | if value of cell 3 at time (18 + 1) = value of cell 3 at time 19 and value of cell 3 at time (18 + 1) = value of cell 3 at time 18, then value of cell 3 at time 19 = value of cell 3 at time 18 | 
| 6 | value of cell 3 at time 19 = 3 | if value of cell 3 at time 19 = value of cell 3 at time 18 and value of cell 3 at time 18 = 3, then value of cell 3 at time 19 = 3 | 
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