Proof: Get Begin Expr Params Unchanged3
Let's prove the following theorem:
if the following are true:
- the expression at time 3 = 1
- expression state at time 3 = "begin_expr"
- 1 is constant
- arguments stack at time 3 = [ [ 3, [ ] ], [ ] ]
then arguments stack at time 4 = [ [ 3, [ ] ], [ ] ]
Proof:
Given
1 | the expression at time 3 = 1 |
---|---|
2 | expression state at time 3 = "begin_expr" |
3 | 1 is constant |
4 | arguments stack at time 3 = [ [ 3, [ ] ], [ ] ] |
# | Claim | Reason |
---|---|---|
1 | arguments stack at time (3 + 1) = arguments stack at time 3 | if expression state at time 3 = "begin_expr" and the expression at time 3 = 1 and 1 is constant, then arguments stack at time (3 + 1) = arguments stack at time 3 |
2 | arguments stack at time (3 + 1) = [ [ 3, [ ] ], [ ] ] | if arguments stack at time (3 + 1) = arguments stack at time 3 and arguments stack at time 3 = [ [ 3, [ ] ], [ ] ], then arguments stack at time (3 + 1) = [ [ 3, [ ] ], [ ] ] |
3 | 3 + 1 = 4 | 3 + 1 = 4 |
4 | arguments stack at time (3 + 1) = arguments stack at time 4 | if 3 + 1 = 4, then arguments stack at time (3 + 1) = arguments stack at time 4 |
5 | arguments stack at time 4 = [ [ 3, [ ] ], [ ] ] | if arguments stack at time (3 + 1) = arguments stack at time 4 and arguments stack at time (3 + 1) = [ [ 3, [ ] ], [ ] ], then arguments stack at time 4 = [ [ 3, [ ] ], [ ] ] |
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