Proof: Get Begin Expr Params 12

Let's prove the following theorem:

if the following are true:

expression state at time 12 = "begin_expr"
the expression at time 12 = Dog()
arguments stack at time 12 = [ ]

then arguments stack at time 13 = [ [ ], [ ] ]

Proof:

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Given

1	expression state at time 12 = "begin_expr"
2	the expression at time 12 = `Dog()`
3	arguments stack at time 12 = [ ]

Proof Table

#	Claim	Reason
1	arguments stack at time (12 + 1) = [ [ ], arguments stack at time 12 ]	if expression state at time 12 = "begin_expr" and the expression at time 12 = `Dog()`, then arguments stack at time (12 + 1) = [ [ ], arguments stack at time 12 ] (Begin Constructor Arguments Property)
2	[ [ ], arguments stack at time 12 ] = [ [ ], [ ] ]	if arguments stack at time 12 = [ ], then [ [ ], arguments stack at time 12 ] = [ [ ], [ ] ] (Substitution)
3	arguments stack at time (12 + 1) = [ [ ], [ ] ]	if arguments stack at time (12 + 1) = [ [ ], arguments stack at time 12 ] and [ [ ], arguments stack at time 12 ] = [ [ ], [ ] ], then arguments stack at time (12 + 1) = [ [ ], [ ] ] (Transitive Property of Equality)
4	12 + 1 = 13	12 + 1 = 13 (Fixed Value Statement)
5	arguments stack at time (12 + 1) = arguments stack at time 13	if 12 + 1 = 13, then arguments stack at time (12 + 1) = arguments stack at time 13 (Substitution)
6	arguments stack at time 13 = [ [ ], [ ] ]	if arguments stack at time (12 + 1) = arguments stack at time 13 and arguments stack at time (12 + 1) = [ [ ], [ ] ], then arguments stack at time 13 = [ [ ], [ ] ] (Transitive Property of Equality Variation 2)

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