Proof: Pc 3
Let's prove the following theorem:
if the following are true:
- subtract immediate instruction with dst: 1 src: 1 and immediate: 1 at 9
- the PC at time 3 = 9
then the PC at time 4 = 10
Proof:
Given
1 | subtract immediate instruction with dst: 1 src: 1 and immediate: 1 at 9 |
---|---|
2 | the PC at time 3 = 9 |
# | Claim | Reason |
---|---|---|
1 | the PC at time (3 + 1) = 9 + 1 | if subtract immediate instruction with dst: 1 src: 1 and immediate: 1 at 9 and the PC at time 3 = 9, then the PC at time (3 + 1) = 9 + 1 |
2 | 3 + 1 = 4 | 3 + 1 = 4 |
3 | 9 + 1 = 10 | 9 + 1 = 10 |
4 | the PC at time (3 + 1) = the PC at time 4 | if 3 + 1 = 4, then the PC at time (3 + 1) = the PC at time 4 |
5 | the PC at time 4 = 10 | if the PC at time (3 + 1) = 9 + 1 and the PC at time (3 + 1) = the PC at time 4 and 9 + 1 = 10, then the PC at time 4 = 10 |
Comments
Please log in to add comments