Proof: Byte 8 Stays the Same 7
Let's prove the following theorem:
if the following are true:
- instruction #0 is
load dst=7 addr=5 imm=0
- the PC at time 7 = 0
- value of cell 8 at time 7 = 24
then value of cell 8 at time 8 = 24
Proof:
Given
1 | instruction #0 is load dst=7 addr=5 imm=0 |
---|---|
2 | the PC at time 7 = 0 |
3 | value of cell 8 at time 7 = 24 |
# | Claim | Reason |
---|---|---|
1 | not (8 = 7) | not (8 = 7) |
2 | value of cell 8 at time (7 + 1) = value of cell 8 at time 7 | if instruction #0 is load dst=7 addr=5 imm=0 and the PC at time 7 = 0 and not (8 = 7), then value of cell 8 at time (7 + 1) = value of cell 8 at time 7 |
3 | 7 + 1 = 8 | 7 + 1 = 8 |
4 | value of cell 8 at time (7 + 1) = value of cell 8 at time 8 | if 7 + 1 = 8, then value of cell 8 at time (7 + 1) = value of cell 8 at time 8 |
5 | value of cell 8 at time 8 = value of cell 8 at time 7 | if value of cell 8 at time (7 + 1) = value of cell 8 at time 8 and value of cell 8 at time (7 + 1) = value of cell 8 at time 7, then value of cell 8 at time 8 = value of cell 8 at time 7 |
6 | value of cell 8 at time 8 = 24 | if value of cell 8 at time 8 = value of cell 8 at time 7 and value of cell 8 at time 7 = 24, then value of cell 8 at time 8 = 24 |
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