Proof: Get Begin Expr Params Unchanged36

Let's prove the following theorem:

if the following are true:

the expression at time 36 = "Spain"
expression state at time 36 = "begin_expr"
"Spain" is constant
arguments stack at time 36 = [ [ ], [ ] ]

then arguments stack at time 37 = [ [ ], [ ] ]

Proof:

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Given

1	the expression at time 36 = "Spain"
2	expression state at time 36 = "begin_expr"
3	"Spain" is constant
4	arguments stack at time 36 = [ [ ], [ ] ]

Proof Table

#	Claim	Reason
1	arguments stack at time (36 + 1) = arguments stack at time 36	if expression state at time 36 = "begin_expr" and the expression at time 36 = "Spain" and "Spain" is constant, then arguments stack at time (36 + 1) = arguments stack at time 36 (Evaluate Constant (3))
2	arguments stack at time (36 + 1) = [ [ ], [ ] ]	if arguments stack at time (36 + 1) = arguments stack at time 36 and arguments stack at time 36 = [ [ ], [ ] ], then arguments stack at time (36 + 1) = [ [ ], [ ] ] (Transitive Property of Equality)
3	36 + 1 = 37	36 + 1 = 37 (Fixed Value Statement)
4	arguments stack at time (36 + 1) = arguments stack at time 37	if 36 + 1 = 37, then arguments stack at time (36 + 1) = arguments stack at time 37 (Substitution)
5	arguments stack at time 37 = [ [ ], [ ] ]	if arguments stack at time (36 + 1) = arguments stack at time 37 and arguments stack at time (36 + 1) = [ [ ], [ ] ], then arguments stack at time 37 = [ [ ], [ ] ] (Transitive Property of Equality Variation 2)

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