Proof: Pc 14
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 14 = 0
then the PC at time 15 = 1
Proof:
Given
| 1 | instruction #0 is load dst=7 addr=5 imm=0 |
|---|---|
| 2 | the PC at time 14 = 0 |
| # | Claim | Reason |
|---|---|---|
| 1 | the PC at time (14 + 1) = 0 + 1 | if instruction #0 is load dst=7 addr=5 imm=0 and the PC at time 14 = 0, then the PC at time (14 + 1) = 0 + 1 |
| 2 | 14 + 1 = 15 | 14 + 1 = 15 |
| 3 | 0 + 1 = 1 | 0 + 1 = 1 |
| 4 | the PC at time (14 + 1) = the PC at time 15 | if 14 + 1 = 15, then the PC at time (14 + 1) = the PC at time 15 |
| 5 | the PC at time 15 = 1 | if the PC at time (14 + 1) = 0 + 1 and the PC at time (14 + 1) = the PC at time 15 and 0 + 1 = 1, then the PC at time 15 = 1 |
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