Proof: Get Begin Expr Params Unchanged19

Let's prove the following theorem:

if the following are true:

the expression at time 19 = "Mexico"
expression state at time 19 = "begin_expr"
"Mexico" is constant
arguments stack at time 19 = [ [ ], [ [ ], [ ] ] ]

then arguments stack at time 20 = [ [ ], [ [ ], [ ] ] ]

Proof:

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Given

1	the expression at time 19 = "Mexico"
2	expression state at time 19 = "begin_expr"
3	"Mexico" is constant
4	arguments stack at time 19 = [ [ ], [ [ ], [ ] ] ]

Proof Table

#	Claim	Reason
1	arguments stack at time (19 + 1) = arguments stack at time 19	if expression state at time 19 = "begin_expr" and the expression at time 19 = "Mexico" and "Mexico" is constant, then arguments stack at time (19 + 1) = arguments stack at time 19 (Evaluate Constant (3))
2	arguments stack at time (19 + 1) = [ [ ], [ [ ], [ ] ] ]	if arguments stack at time (19 + 1) = arguments stack at time 19 and arguments stack at time 19 = [ [ ], [ [ ], [ ] ] ], then arguments stack at time (19 + 1) = [ [ ], [ [ ], [ ] ] ] (Transitive Property of Equality)
3	19 + 1 = 20	19 + 1 = 20 (Fixed Value Statement)
4	arguments stack at time (19 + 1) = arguments stack at time 20	if 19 + 1 = 20, then arguments stack at time (19 + 1) = arguments stack at time 20 (Substitution)
5	arguments stack at time 20 = [ [ ], [ [ ], [ ] ] ]	if arguments stack at time (19 + 1) = arguments stack at time 20 and arguments stack at time (19 + 1) = [ [ ], [ [ ], [ ] ] ], then arguments stack at time 20 = [ [ ], [ [ ], [ ] ] ] (Transitive Property of Equality Variation 2)

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