Proof: Byte 5 Stays the Same 0
Let's prove the following theorem:
if the following are true:
- instruction #0 is
load dst=7 addr=6 imm=0 - the PC at time 0 = 0
- value of cell 5 at time 0 = 20
then value of cell 5 at time 1 = 20
Proof:
Given
| 1 | instruction #0 is load dst=7 addr=6 imm=0 |
|---|---|
| 2 | the PC at time 0 = 0 |
| 3 | value of cell 5 at time 0 = 20 |
| # | Claim | Reason |
|---|---|---|
| 1 | not (5 = 7) | not (5 = 7) |
| 2 | value of cell 5 at time (0 + 1) = value of cell 5 at time 0 | if instruction #0 is load dst=7 addr=6 imm=0 and the PC at time 0 = 0 and not (5 = 7), then value of cell 5 at time (0 + 1) = value of cell 5 at time 0 |
| 3 | 0 + 1 = 1 | 0 + 1 = 1 |
| 4 | value of cell 5 at time (0 + 1) = value of cell 5 at time 1 | if 0 + 1 = 1, then value of cell 5 at time (0 + 1) = value of cell 5 at time 1 |
| 5 | value of cell 5 at time 1 = value of cell 5 at time 0 | if value of cell 5 at time (0 + 1) = value of cell 5 at time 1 and value of cell 5 at time (0 + 1) = value of cell 5 at time 0, then value of cell 5 at time 1 = value of cell 5 at time 0 |
| 6 | value of cell 5 at time 1 = 20 | if value of cell 5 at time 1 = value of cell 5 at time 0 and value of cell 5 at time 0 = 20, then value of cell 5 at time 1 = 20 |
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