Proof: Byte 1 Stays the Same 11
Let's prove the following theorem:
if the following are true:
- instruction #17 is
load dst=3 addr=1 imm=1 - the PC at time 11 = 17
- value of cell 1 at time 11 = 125
then value of cell 1 at time 12 = 125
Proof:
Given
| 1 | instruction #17 is load dst=3 addr=1 imm=1 |
|---|---|
| 2 | the PC at time 11 = 17 |
| 3 | value of cell 1 at time 11 = 125 |
| # | Claim | Reason |
|---|---|---|
| 1 | not (1 = 3) | not (1 = 3) |
| 2 | value of cell 1 at time (11 + 1) = value of cell 1 at time 11 | if instruction #17 is load dst=3 addr=1 imm=1 and the PC at time 11 = 17 and not (1 = 3), then value of cell 1 at time (11 + 1) = value of cell 1 at time 11 |
| 3 | 11 + 1 = 12 | 11 + 1 = 12 |
| 4 | value of cell 1 at time (11 + 1) = value of cell 1 at time 12 | if 11 + 1 = 12, then value of cell 1 at time (11 + 1) = value of cell 1 at time 12 |
| 5 | value of cell 1 at time 12 = value of cell 1 at time 11 | if value of cell 1 at time (11 + 1) = value of cell 1 at time 12 and value of cell 1 at time (11 + 1) = value of cell 1 at time 11, then value of cell 1 at time 12 = value of cell 1 at time 11 |
| 6 | value of cell 1 at time 12 = 125 | if value of cell 1 at time 12 = value of cell 1 at time 11 and value of cell 1 at time 11 = 125, then value of cell 1 at time 12 = 125 |
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