Proof: Get Begin Expr Params Unchanged17
Let's prove the following theorem:
if the following are true:
- the expression at time 17 = 25
- expression state at time 17 = "begin_expr"
- 25 is constant
- arguments stack at time 17 = [ [ ], [ ] ]
then arguments stack at time 18 = [ [ ], [ ] ]
Proof:
Given
1 | the expression at time 17 = 25 |
---|---|
2 | expression state at time 17 = "begin_expr" |
3 | 25 is constant |
4 | arguments stack at time 17 = [ [ ], [ ] ] |
# | Claim | Reason |
---|---|---|
1 | arguments stack at time (17 + 1) = arguments stack at time 17 | if expression state at time 17 = "begin_expr" and the expression at time 17 = 25 and 25 is constant, then arguments stack at time (17 + 1) = arguments stack at time 17 |
2 | arguments stack at time (17 + 1) = [ [ ], [ ] ] | if arguments stack at time (17 + 1) = arguments stack at time 17 and arguments stack at time 17 = [ [ ], [ ] ], then arguments stack at time (17 + 1) = [ [ ], [ ] ] |
3 | 17 + 1 = 18 | 17 + 1 = 18 |
4 | arguments stack at time (17 + 1) = arguments stack at time 18 | if 17 + 1 = 18, then arguments stack at time (17 + 1) = arguments stack at time 18 |
5 | arguments stack at time 18 = [ [ ], [ ] ] | if arguments stack at time (17 + 1) = arguments stack at time 18 and arguments stack at time (17 + 1) = [ [ ], [ ] ], then arguments stack at time 18 = [ [ ], [ ] ] |
Comments
Please log in to add comments