Proof: Get Begin Expr Parent 1

Let's prove the following theorem:

if the following are true:

expression state at time 1 = "begin_expr"
the expression at time 1 = __eq__(1, 3)
parent stack at time 1 = [ ]

then parent stack at time 2 = [ __eq__(1, 3), [ ] ]

Proof:

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Given

1	expression state at time 1 = "begin_expr"
2	the expression at time 1 = `__eq__(1, 3)`
3	parent stack at time 1 = [ ]

Proof Table

#	Claim	Reason
1	parent stack at time (1 + 1) = [ `__eq__(1, 3)`, parent stack at time 1 ]	if expression state at time 1 = "begin_expr" and the expression at time 1 = `__eq__(1, 3)`, then parent stack at time (1 + 1) = [ `__eq__(1, 3)`, parent stack at time 1 ] (Begin Expression Parent Rule)
2	[ `__eq__(1, 3)`, parent stack at time 1 ] = [ `__eq__(1, 3)`, [ ] ]	if parent stack at time 1 = [ ], then [ `__eq__(1, 3)`, parent stack at time 1 ] = [ `__eq__(1, 3)`, [ ] ] (Substitution)
3	parent stack at time (1 + 1) = [ `__eq__(1, 3)`, [ ] ]	if parent stack at time (1 + 1) = [ `__eq__(1, 3)`, parent stack at time 1 ] and [ `__eq__(1, 3)`, parent stack at time 1 ] = [ `__eq__(1, 3)`, [ ] ], then parent stack at time (1 + 1) = [ `__eq__(1, 3)`, [ ] ] (Transitive Property of Equality)
4	1 + 1 = 2	1 + 1 = 2 (Fixed Value Statement)
5	parent stack at time (1 + 1) = parent stack at time 2	if 1 + 1 = 2, then parent stack at time (1 + 1) = parent stack at time 2 (Substitution)
6	parent stack at time 2 = [ `__eq__(1, 3)`, [ ] ]	if parent stack at time (1 + 1) = parent stack at time 2 and parent stack at time (1 + 1) = [ `__eq__(1, 3)`, [ ] ], then parent stack at time 2 = [ `__eq__(1, 3)`, [ ] ] (Transitive Property of Equality Variation 2)

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