Proof: Get Begin Expr Params 47
Let's prove the following theorem:
if the following are true:
- expression state at time 47 = "begin_expr"
- the expression at time 47 =
self.x
- arguments stack at time 47 = [ [ 5, [ ] ], [ [ ], [ ] ] ]
then arguments stack at time 48 = [ [ self, [ ] ], [ [ 5, [ ] ], [ [ ], [ ] ] ] ]
Proof:
Given
1 | expression state at time 47 = "begin_expr" |
---|---|
2 | the expression at time 47 = self.x |
3 | arguments stack at time 47 = [ [ 5, [ ] ], [ [ ], [ ] ] ] |
# | Claim | Reason |
---|---|---|
1 | arguments stack at time (47 + 1) = [ [ self, [ ] ], arguments stack at time 47 ] | if expression state at time 47 = "begin_expr" and the expression at time 47 = self.x , then arguments stack at time (47 + 1) = [ [ self, [ ] ], arguments stack at time 47 ] |
2 | [ [ self, [ ] ], arguments stack at time 47 ] = [ [ self, [ ] ], [ [ 5, [ ] ], [ [ ], [ ] ] ] ] | if arguments stack at time 47 = [ [ 5, [ ] ], [ [ ], [ ] ] ], then [ [ self, [ ] ], arguments stack at time 47 ] = [ [ self, [ ] ], [ [ 5, [ ] ], [ [ ], [ ] ] ] ] |
3 | arguments stack at time (47 + 1) = [ [ self, [ ] ], [ [ 5, [ ] ], [ [ ], [ ] ] ] ] | if arguments stack at time (47 + 1) = [ [ self, [ ] ], arguments stack at time 47 ] and [ [ self, [ ] ], arguments stack at time 47 ] = [ [ self, [ ] ], [ [ 5, [ ] ], [ [ ], [ ] ] ] ], then arguments stack at time (47 + 1) = [ [ self, [ ] ], [ [ 5, [ ] ], [ [ ], [ ] ] ] ] |
4 | 47 + 1 = 48 | 47 + 1 = 48 |
5 | arguments stack at time (47 + 1) = arguments stack at time 48 | if 47 + 1 = 48, then arguments stack at time (47 + 1) = arguments stack at time 48 |
6 | arguments stack at time 48 = [ [ self, [ ] ], [ [ 5, [ ] ], [ [ ], [ ] ] ] ] | if arguments stack at time (47 + 1) = arguments stack at time 48 and arguments stack at time (47 + 1) = [ [ self, [ ] ], [ [ 5, [ ] ], [ [ ], [ ] ] ] ], then arguments stack at time 48 = [ [ self, [ ] ], [ [ 5, [ ] ], [ [ ], [ ] ] ] ] |
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