Proof: Byte 3 Stays the Same 16
Let's prove the following theorem:
if the following are true:
- instruction #3 is
load dst=4 addr=1 imm=0 - the PC at time 16 = 3
- value of cell 3 at time 16 = 2
then value of cell 3 at time 17 = 2
Proof:
Given
| 1 | instruction #3 is load dst=4 addr=1 imm=0 |
|---|---|
| 2 | the PC at time 16 = 3 |
| 3 | value of cell 3 at time 16 = 2 |
| # | Claim | Reason |
|---|---|---|
| 1 | not (3 = 4) | not (3 = 4) |
| 2 | value of cell 3 at time (16 + 1) = value of cell 3 at time 16 | if instruction #3 is load dst=4 addr=1 imm=0 and the PC at time 16 = 3 and not (3 = 4), then value of cell 3 at time (16 + 1) = value of cell 3 at time 16 |
| 3 | 16 + 1 = 17 | 16 + 1 = 17 |
| 4 | value of cell 3 at time (16 + 1) = value of cell 3 at time 17 | if 16 + 1 = 17, then value of cell 3 at time (16 + 1) = value of cell 3 at time 17 |
| 5 | value of cell 3 at time 17 = value of cell 3 at time 16 | if value of cell 3 at time (16 + 1) = value of cell 3 at time 17 and value of cell 3 at time (16 + 1) = value of cell 3 at time 16, then value of cell 3 at time 17 = value of cell 3 at time 16 |
| 6 | value of cell 3 at time 17 = 2 | if value of cell 3 at time 17 = value of cell 3 at time 16 and value of cell 3 at time 16 = 2, then value of cell 3 at time 17 = 2 |
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