Proof: Get Begin Expr Params 18

Let's prove the following theorem:

if the following are true:

expression state at time 18 = "begin_expr"
the expression at time 18 = self."x" = 0
arguments stack at time 18 = [ ]

then arguments stack at time 19 = [ [ 0, [ ] ], [ ] ]

Proof:

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Given

1	expression state at time 18 = "begin_expr"
2	the expression at time 18 = self."x" = 0
3	arguments stack at time 18 = [ ]

Proof Table

#	Claim	Reason
1	arguments stack at time (18 + 1) = [ [ 0, [ ] ], arguments stack at time 18 ]	if expression state at time 18 = "begin_expr" and the expression at time 18 = self."x" = 0, then arguments stack at time (18 + 1) = [ [ 0, [ ] ], arguments stack at time 18 ] (Set Attribute Begin Arguments Property)
2	[ [ 0, [ ] ], arguments stack at time 18 ] = [ [ 0, [ ] ], [ ] ]	if arguments stack at time 18 = [ ], then [ [ 0, [ ] ], arguments stack at time 18 ] = [ [ 0, [ ] ], [ ] ] (Substitution)
3	arguments stack at time (18 + 1) = [ [ 0, [ ] ], [ ] ]	if arguments stack at time (18 + 1) = [ [ 0, [ ] ], arguments stack at time 18 ] and [ [ 0, [ ] ], arguments stack at time 18 ] = [ [ 0, [ ] ], [ ] ], then arguments stack at time (18 + 1) = [ [ 0, [ ] ], [ ] ] (Transitive Property of Equality)
4	18 + 1 = 19	18 + 1 = 19 (Fixed Value Statement)
5	arguments stack at time (18 + 1) = arguments stack at time 19	if 18 + 1 = 19, then arguments stack at time (18 + 1) = arguments stack at time 19 (Substitution)
6	arguments stack at time 19 = [ [ 0, [ ] ], [ ] ]	if arguments stack at time (18 + 1) = arguments stack at time 19 and arguments stack at time (18 + 1) = [ [ 0, [ ] ], [ ] ], then arguments stack at time 19 = [ [ 0, [ ] ], [ ] ] (Transitive Property of Equality Variation 2)

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