Proof: Subtracti Insn 6
Let's prove the following theorem:
if the following are true:
- subtract immediate instruction with dst: 1 src: 1 and immediate: 1 at 12
- the PC at time 6 = 12
- value of cell 1 at time 6 = 127
then value of cell 1 at time 7 = 126
Proof:
Given
1 | subtract immediate instruction with dst: 1 src: 1 and immediate: 1 at 12 |
---|---|
2 | the PC at time 6 = 12 |
3 | value of cell 1 at time 6 = 127 |
# | Claim | Reason |
---|---|---|
1 | value of cell 1 at time (6 + 1) = (value of cell 1 at time 6) - 1 | if subtract immediate instruction with dst: 1 src: 1 and immediate: 1 at 12 and the PC at time 6 = 12, then value of cell 1 at time (6 + 1) = (value of cell 1 at time 6) - 1 |
2 | 6 + 1 = 7 | 6 + 1 = 7 |
3 | value of cell 1 at time (6 + 1) = value of cell 1 at time 7 | if 6 + 1 = 7, then value of cell 1 at time (6 + 1) = value of cell 1 at time 7 |
4 | value of cell 1 at time 7 = (value of cell 1 at time 6) - 1 | if value of cell 1 at time (6 + 1) = value of cell 1 at time 7 and value of cell 1 at time (6 + 1) = (value of cell 1 at time 6) - 1, then value of cell 1 at time 7 = (value of cell 1 at time 6) - 1 |
5 | (value of cell 1 at time 6) - 1 = 127 - 1 | if value of cell 1 at time 6 = 127, then (value of cell 1 at time 6) - 1 = 127 - 1 |
6 | value of cell 1 at time 7 = 127 - 1 | if value of cell 1 at time 7 = (value of cell 1 at time 6) - 1 and (value of cell 1 at time 6) - 1 = 127 - 1, then value of cell 1 at time 7 = 127 - 1 |
7 | 127 - 1 = 126 | 127 - 1 = 126 |
8 | value of cell 1 at time 7 = 126 | if value of cell 1 at time 7 = 127 - 1 and 127 - 1 = 126, then value of cell 1 at time 7 = 126 |
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