Proof: Get Begin Expr Params Unchanged11
Let's prove the following theorem:
if the following are true:
- the expression at time 11 = "John"
- expression state at time 11 = "begin_expr"
- "John" is constant
- arguments stack at time 11 = [ [ "Smith", [ 25, [ ] ] ], [ ] ]
then arguments stack at time 12 = [ [ "Smith", [ 25, [ ] ] ], [ ] ]
Proof:
Given
1 | the expression at time 11 = "John" |
---|---|
2 | expression state at time 11 = "begin_expr" |
3 | "John" is constant |
4 | arguments stack at time 11 = [ [ "Smith", [ 25, [ ] ] ], [ ] ] |
# | Claim | Reason |
---|---|---|
1 | arguments stack at time (11 + 1) = arguments stack at time 11 | if expression state at time 11 = "begin_expr" and the expression at time 11 = "John" and "John" is constant, then arguments stack at time (11 + 1) = arguments stack at time 11 |
2 | arguments stack at time (11 + 1) = [ [ "Smith", [ 25, [ ] ] ], [ ] ] | if arguments stack at time (11 + 1) = arguments stack at time 11 and arguments stack at time 11 = [ [ "Smith", [ 25, [ ] ] ], [ ] ], then arguments stack at time (11 + 1) = [ [ "Smith", [ 25, [ ] ] ], [ ] ] |
3 | 11 + 1 = 12 | 11 + 1 = 12 |
4 | arguments stack at time (11 + 1) = arguments stack at time 12 | if 11 + 1 = 12, then arguments stack at time (11 + 1) = arguments stack at time 12 |
5 | arguments stack at time 12 = [ [ "Smith", [ 25, [ ] ] ], [ ] ] | if arguments stack at time (11 + 1) = arguments stack at time 12 and arguments stack at time (11 + 1) = [ [ "Smith", [ 25, [ ] ] ], [ ] ], then arguments stack at time 12 = [ [ "Smith", [ 25, [ ] ] ], [ ] ] |
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