Proof: Get Begin Expr Params Unchanged55
Let's prove the following theorem:
if the following are true:
- the expression at time 55 = 5
- expression state at time 55 = "begin_expr"
- 5 is constant
- arguments stack at time 55 = [ [ ], [ [ ], [ ] ] ]
then arguments stack at time 56 = [ [ ], [ [ ], [ ] ] ]
Proof:
Given
1 | the expression at time 55 = 5 |
---|---|
2 | expression state at time 55 = "begin_expr" |
3 | 5 is constant |
4 | arguments stack at time 55 = [ [ ], [ [ ], [ ] ] ] |
# | Claim | Reason |
---|---|---|
1 | arguments stack at time (55 + 1) = arguments stack at time 55 | if expression state at time 55 = "begin_expr" and the expression at time 55 = 5 and 5 is constant, then arguments stack at time (55 + 1) = arguments stack at time 55 |
2 | arguments stack at time (55 + 1) = [ [ ], [ [ ], [ ] ] ] | if arguments stack at time (55 + 1) = arguments stack at time 55 and arguments stack at time 55 = [ [ ], [ [ ], [ ] ] ], then arguments stack at time (55 + 1) = [ [ ], [ [ ], [ ] ] ] |
3 | 55 + 1 = 56 | 55 + 1 = 56 |
4 | arguments stack at time (55 + 1) = arguments stack at time 56 | if 55 + 1 = 56, then arguments stack at time (55 + 1) = arguments stack at time 56 |
5 | arguments stack at time 56 = [ [ ], [ [ ], [ ] ] ] | if arguments stack at time (55 + 1) = arguments stack at time 56 and arguments stack at time (55 + 1) = [ [ ], [ [ ], [ ] ] ], then arguments stack at time 56 = [ [ ], [ [ ], [ ] ] ] |
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