Proof: Get Begin Expr Params 35

Let's prove the following theorem:

if the following are true:

expression state at time 35 = "begin_expr"
the expression at time 35 = self."last_name" = last_name
arguments stack at time 35 = [ ]

then arguments stack at time 36 = [ [ last_name, [ ] ], [ ] ]

Proof:

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Given

1	expression state at time 35 = "begin_expr"
2	the expression at time 35 = self."last_name" = last_name
3	arguments stack at time 35 = [ ]

Proof Table

#	Claim	Reason
1	arguments stack at time (35 + 1) = [ [ last_name, [ ] ], arguments stack at time 35 ]	if expression state at time 35 = "begin_expr" and the expression at time 35 = self."last_name" = last_name, then arguments stack at time (35 + 1) = [ [ last_name, [ ] ], arguments stack at time 35 ] (Set Attribute Begin Arguments Property)
2	[ [ last_name, [ ] ], arguments stack at time 35 ] = [ [ last_name, [ ] ], [ ] ]	if arguments stack at time 35 = [ ], then [ [ last_name, [ ] ], arguments stack at time 35 ] = [ [ last_name, [ ] ], [ ] ] (Substitution)
3	arguments stack at time (35 + 1) = [ [ last_name, [ ] ], [ ] ]	if arguments stack at time (35 + 1) = [ [ last_name, [ ] ], arguments stack at time 35 ] and [ [ last_name, [ ] ], arguments stack at time 35 ] = [ [ last_name, [ ] ], [ ] ], then arguments stack at time (35 + 1) = [ [ last_name, [ ] ], [ ] ] (Transitive Property of Equality)
4	35 + 1 = 36	35 + 1 = 36 (Fixed Value Statement)
5	arguments stack at time (35 + 1) = arguments stack at time 36	if 35 + 1 = 36, then arguments stack at time (35 + 1) = arguments stack at time 36 (Substitution)
6	arguments stack at time 36 = [ [ last_name, [ ] ], [ ] ]	if arguments stack at time (35 + 1) = arguments stack at time 36 and arguments stack at time (35 + 1) = [ [ last_name, [ ] ], [ ] ], then arguments stack at time 36 = [ [ last_name, [ ] ], [ ] ] (Transitive Property of Equality Variation 2)

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