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