Proof: Get Begin Expr Params Unchanged16

Let's prove the following theorem:

if the following are true:

the expression at time 16 = "MX"
expression state at time 16 = "begin_expr"
"MX" is constant
arguments stack at time 16 = [ [ "Mexico", [ ] ], [ [ ], [ ] ] ]

then arguments stack at time 17 = [ [ "Mexico", [ ] ], [ [ ], [ ] ] ]

Proof:

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Given

1	the expression at time 16 = "MX"
2	expression state at time 16 = "begin_expr"
3	"MX" is constant
4	arguments stack at time 16 = [ [ "Mexico", [ ] ], [ [ ], [ ] ] ]

Proof Table

#	Claim	Reason
1	arguments stack at time (16 + 1) = arguments stack at time 16	if expression state at time 16 = "begin_expr" and the expression at time 16 = "MX" and "MX" is constant, then arguments stack at time (16 + 1) = arguments stack at time 16 (Evaluate Constant (3))
2	arguments stack at time (16 + 1) = [ [ "Mexico", [ ] ], [ [ ], [ ] ] ]	if arguments stack at time (16 + 1) = arguments stack at time 16 and arguments stack at time 16 = [ [ "Mexico", [ ] ], [ [ ], [ ] ] ], then arguments stack at time (16 + 1) = [ [ "Mexico", [ ] ], [ [ ], [ ] ] ] (Transitive Property of Equality)
3	16 + 1 = 17	16 + 1 = 17 (Fixed Value Statement)
4	arguments stack at time (16 + 1) = arguments stack at time 17	if 16 + 1 = 17, then arguments stack at time (16 + 1) = arguments stack at time 17 (Substitution)
5	arguments stack at time 17 = [ [ "Mexico", [ ] ], [ [ ], [ ] ] ]	if arguments stack at time (16 + 1) = arguments stack at time 17 and arguments stack at time (16 + 1) = [ [ "Mexico", [ ] ], [ [ ], [ ] ] ], then arguments stack at time 17 = [ [ "Mexico", [ ] ], [ [ ], [ ] ] ] (Transitive Property of Equality Variation 2)

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