Proof: Get Begin Expr Params 14

Let's prove the following theorem:

if the following are true:
  • expression state at time 14 = "begin_expr"
  • the expression at time 14 = entry "MX": "Mexico"
  • arguments stack at time 14 = [ [ ], [ ] ]

then arguments stack at time 15 = [ [ "MX", [ "Mexico", [ ] ] ], [ [ ], [ ] ] ]

Proof:

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Given
1 expression state at time 14 = "begin_expr"
2 the expression at time 14 = entry "MX": "Mexico"
3 arguments stack at time 14 = [ [ ], [ ] ]
Proof Table
# Claim Reason
1 arguments stack at time (14 + 1) = [ [ "MX", [ "Mexico", [ ] ] ], arguments stack at time 14 ] if expression state at time 14 = "begin_expr" and the expression at time 14 = entry "MX": "Mexico", then arguments stack at time (14 + 1) = [ [ "MX", [ "Mexico", [ ] ] ], arguments stack at time 14 ]
2 [ [ "MX", [ "Mexico", [ ] ] ], arguments stack at time 14 ] = [ [ "MX", [ "Mexico", [ ] ] ], [ [ ], [ ] ] ] if arguments stack at time 14 = [ [ ], [ ] ], then [ [ "MX", [ "Mexico", [ ] ] ], arguments stack at time 14 ] = [ [ "MX", [ "Mexico", [ ] ] ], [ [ ], [ ] ] ]
3 arguments stack at time (14 + 1) = [ [ "MX", [ "Mexico", [ ] ] ], [ [ ], [ ] ] ] if arguments stack at time (14 + 1) = [ [ "MX", [ "Mexico", [ ] ] ], arguments stack at time 14 ] and [ [ "MX", [ "Mexico", [ ] ] ], arguments stack at time 14 ] = [ [ "MX", [ "Mexico", [ ] ] ], [ [ ], [ ] ] ], then arguments stack at time (14 + 1) = [ [ "MX", [ "Mexico", [ ] ] ], [ [ ], [ ] ] ]
4 14 + 1 = 15 14 + 1 = 15
5 arguments stack at time (14 + 1) = arguments stack at time 15 if 14 + 1 = 15, then arguments stack at time (14 + 1) = arguments stack at time 15
6 arguments stack at time 15 = [ [ "MX", [ "Mexico", [ ] ] ], [ [ ], [ ] ] ] if arguments stack at time (14 + 1) = arguments stack at time 15 and arguments stack at time (14 + 1) = [ [ "MX", [ "Mexico", [ ] ] ], [ [ ], [ ] ] ], then arguments stack at time 15 = [ [ "MX", [ "Mexico", [ ] ] ], [ [ ], [ ] ] ]

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