Proof: Get Begin Expr Params Unchanged33

Let's prove the following theorem:

if the following are true:

the expression at time 33 = "ES"
expression state at time 33 = "begin_expr"
"ES" is constant
arguments stack at time 33 = [ [ "Spain", [ ] ], [ ] ]

then arguments stack at time 34 = [ [ "Spain", [ ] ], [ ] ]

Proof:

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Given

1	the expression at time 33 = "ES"
2	expression state at time 33 = "begin_expr"
3	"ES" is constant
4	arguments stack at time 33 = [ [ "Spain", [ ] ], [ ] ]

Proof Table

#	Claim	Reason
1	arguments stack at time (33 + 1) = arguments stack at time 33	if expression state at time 33 = "begin_expr" and the expression at time 33 = "ES" and "ES" is constant, then arguments stack at time (33 + 1) = arguments stack at time 33 (Evaluate Constant (3))
2	arguments stack at time (33 + 1) = [ [ "Spain", [ ] ], [ ] ]	if arguments stack at time (33 + 1) = arguments stack at time 33 and arguments stack at time 33 = [ [ "Spain", [ ] ], [ ] ], then arguments stack at time (33 + 1) = [ [ "Spain", [ ] ], [ ] ] (Transitive Property of Equality)
3	33 + 1 = 34	33 + 1 = 34 (Fixed Value Statement)
4	arguments stack at time (33 + 1) = arguments stack at time 34	if 33 + 1 = 34, then arguments stack at time (33 + 1) = arguments stack at time 34 (Substitution)
5	arguments stack at time 34 = [ [ "Spain", [ ] ], [ ] ]	if arguments stack at time (33 + 1) = arguments stack at time 34 and arguments stack at time (33 + 1) = [ [ "Spain", [ ] ], [ ] ], then arguments stack at time 34 = [ [ "Spain", [ ] ], [ ] ] (Transitive Property of Equality Variation 2)

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