Proof: Store 1
Let's prove the following theorem:
if the following are true:
- instruction #1 is
store src=4 addr=6 imm=1
- the PC at time 1 = 1
- value of cell 6 at time 1 = 7
- value of cell 4 at time 1 = 11
then value of cell 8 at time 2 = 11
Proof:
Given
1 | instruction #1 is store src=4 addr=6 imm=1 |
---|---|
2 | the PC at time 1 = 1 |
3 | value of cell 6 at time 1 = 7 |
4 | value of cell 4 at time 1 = 11 |
# | Claim | Reason |
---|---|---|
1 | value of cell ((value of cell 6 at time 1) + 1) at time (1 + 1) = value of cell 4 at time 1 | if instruction #1 is store src=4 addr=6 imm=1 and the PC at time 1 = 1, then value of cell ((value of cell 6 at time 1) + 1) at time (1 + 1) = value of cell 4 at time 1 |
2 | (value of cell 6 at time 1) + 1 = 7 + 1 | if value of cell 6 at time 1 = 7, then (value of cell 6 at time 1) + 1 = 7 + 1 |
3 | 7 + 1 = 8 | 7 + 1 = 8 |
4 | (value of cell 6 at time 1) + 1 = 8 | if (value of cell 6 at time 1) + 1 = 7 + 1 and 7 + 1 = 8, then (value of cell 6 at time 1) + 1 = 8 |
5 | value of cell ((value of cell 6 at time 1) + 1) at time (1 + 1) = value of cell 8 at time (1 + 1) | if (value of cell 6 at time 1) + 1 = 8, then value of cell ((value of cell 6 at time 1) + 1) at time (1 + 1) = value of cell 8 at time (1 + 1) |
6 | 1 + 1 = 2 | 1 + 1 = 2 |
7 | value of cell 8 at time (1 + 1) = value of cell 8 at time 2 | if 1 + 1 = 2, then value of cell 8 at time (1 + 1) = value of cell 8 at time 2 |
8 | value of cell 8 at time 2 = value of cell 4 at time 1 | if value of cell ((value of cell 6 at time 1) + 1) at time (1 + 1) = value of cell 8 at time (1 + 1) and value of cell 8 at time (1 + 1) = value of cell 8 at time 2 and value of cell ((value of cell 6 at time 1) + 1) at time (1 + 1) = value of cell 4 at time 1, then value of cell 8 at time 2 = value of cell 4 at time 1 |
9 | value of cell 8 at time 2 = 11 | if value of cell 8 at time 2 = value of cell 4 at time 1 and value of cell 4 at time 1 = 11, then value of cell 8 at time 2 = 11 |
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