Proof: Byte 127 Stays the Same 9
Let's prove the following theorem:
if the following are true:
    
    
    
    - subtract immediate instruction with dst: 1 src: 1 and immediate: 1 at 15
- the PC at time 9 = 15
- value of cell 127 at time 9 = 2
then value of cell 127 at time 10 = 2
Proof:
  
      
      Given
      
    
    
      
  
  
| 1 | subtract immediate instruction with dst: 1 src: 1 and immediate: 1 at 15 | 
|---|---|
| 2 | the PC at time 9 = 15 | 
| 3 | value of cell 127 at time 9 = 2 | 
| # | Claim | Reason | 
|---|---|---|
| 1 | not (127 = 1) | not (127 = 1) | 
| 2 | value of cell 127 at time (9 + 1) = value of cell 127 at time 9 | if subtract immediate instruction with dst: 1 src: 1 and immediate: 1 at 15 and the PC at time 9 = 15 and not (127 = 1), then value of cell 127 at time (9 + 1) = value of cell 127 at time 9 | 
| 3 | 9 + 1 = 10 | 9 + 1 = 10 | 
| 4 | value of cell 127 at time (9 + 1) = value of cell 127 at time 10 | if 9 + 1 = 10, then value of cell 127 at time (9 + 1) = value of cell 127 at time 10 | 
| 5 | value of cell 127 at time 10 = value of cell 127 at time 9 | if value of cell 127 at time (9 + 1) = value of cell 127 at time 10 and value of cell 127 at time (9 + 1) = value of cell 127 at time 9, then value of cell 127 at time 10 = value of cell 127 at time 9 | 
| 6 | value of cell 127 at time 10 = 2 | if value of cell 127 at time 10 = value of cell 127 at time 9 and value of cell 127 at time 9 = 2, then value of cell 127 at time 10 = 2 | 
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