Proof: Byte 1 Stays the Same 8
Let's prove the following theorem:
if the following are true:
- instruction #14 is
load dst=3 addr=1 imm=1 - the PC at time 8 = 14
- value of cell 1 at time 8 = 126
then value of cell 1 at time 9 = 126
Proof:
Given
| 1 | instruction #14 is load dst=3 addr=1 imm=1 |
|---|---|
| 2 | the PC at time 8 = 14 |
| 3 | value of cell 1 at time 8 = 126 |
| # | Claim | Reason |
|---|---|---|
| 1 | not (1 = 3) | not (1 = 3) |
| 2 | value of cell 1 at time (8 + 1) = value of cell 1 at time 8 | if instruction #14 is load dst=3 addr=1 imm=1 and the PC at time 8 = 14 and not (1 = 3), then value of cell 1 at time (8 + 1) = value of cell 1 at time 8 |
| 3 | 8 + 1 = 9 | 8 + 1 = 9 |
| 4 | value of cell 1 at time (8 + 1) = value of cell 1 at time 9 | if 8 + 1 = 9, then value of cell 1 at time (8 + 1) = value of cell 1 at time 9 |
| 5 | value of cell 1 at time 9 = value of cell 1 at time 8 | if value of cell 1 at time (8 + 1) = value of cell 1 at time 9 and value of cell 1 at time (8 + 1) = value of cell 1 at time 8, then value of cell 1 at time 9 = value of cell 1 at time 8 |
| 6 | value of cell 1 at time 9 = 126 | if value of cell 1 at time 9 = value of cell 1 at time 8 and value of cell 1 at time 8 = 126, then value of cell 1 at time 9 = 126 |
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