Proof: Get Begin Expr Params Unchanged5

Let's prove the following theorem:

if the following are true:
  • the expression at time 5 = "ES"
  • expression state at time 5 = "begin_expr"
  • "ES" is constant
  • arguments stack at time 5 = [ [ "Estonia", [ ] ], [ [ entry "MX": "Mexico", [ ] ], [ ] ] ]

then arguments stack at time 6 = [ [ "Estonia", [ ] ], [ [ entry "MX": "Mexico", [ ] ], [ ] ] ]

Proof:

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Given
1 the expression at time 5 = "ES"
2 expression state at time 5 = "begin_expr"
3 "ES" is constant
4 arguments stack at time 5 = [ [ "Estonia", [ ] ], [ [ entry "MX": "Mexico", [ ] ], [ ] ] ]
Proof Table
# Claim Reason
1 arguments stack at time (5 + 1) = arguments stack at time 5 if expression state at time 5 = "begin_expr" and the expression at time 5 = "ES" and "ES" is constant, then arguments stack at time (5 + 1) = arguments stack at time 5
2 arguments stack at time (5 + 1) = [ [ "Estonia", [ ] ], [ [ entry "MX": "Mexico", [ ] ], [ ] ] ] if arguments stack at time (5 + 1) = arguments stack at time 5 and arguments stack at time 5 = [ [ "Estonia", [ ] ], [ [ entry "MX": "Mexico", [ ] ], [ ] ] ], then arguments stack at time (5 + 1) = [ [ "Estonia", [ ] ], [ [ entry "MX": "Mexico", [ ] ], [ ] ] ]
3 5 + 1 = 6 5 + 1 = 6
4 arguments stack at time (5 + 1) = arguments stack at time 6 if 5 + 1 = 6, then arguments stack at time (5 + 1) = arguments stack at time 6
5 arguments stack at time 6 = [ [ "Estonia", [ ] ], [ [ entry "MX": "Mexico", [ ] ], [ ] ] ] if arguments stack at time (5 + 1) = arguments stack at time 6 and arguments stack at time (5 + 1) = [ [ "Estonia", [ ] ], [ [ entry "MX": "Mexico", [ ] ], [ ] ] ], then arguments stack at time 6 = [ [ "Estonia", [ ] ], [ [ entry "MX": "Mexico", [ ] ], [ ] ] ]

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