Proof: Do Arg Stack At Unchanged 19

Let's prove the following theorem:

if the following are true:

expression state at time 19 = "return"
arguments stack at time 19 = [ [ ], [ ] ]

then arguments stack at time 20 = [ [ ], [ ] ]

Proof:

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Given

1	expression state at time 19 = "return"
2	arguments stack at time 19 = [ [ ], [ ] ]

Proof Table

#	Claim	Reason
1	arguments stack at time (19 + 1) = arguments stack at time 19	if expression state at time 19 = "return", then arguments stack at time (19 + 1) = arguments stack at time 19 (Return (3))
2	arguments stack at time (19 + 1) = [ [ ], [ ] ]	if arguments stack at time (19 + 1) = arguments stack at time 19 and arguments stack at time 19 = [ [ ], [ ] ], then arguments stack at time (19 + 1) = [ [ ], [ ] ] (Transitive Property of Equality)
3	19 + 1 = 20	19 + 1 = 20 (Fixed Value Statement)
4	arguments stack at time (19 + 1) = arguments stack at time 20	if 19 + 1 = 20, then arguments stack at time (19 + 1) = arguments stack at time 20 (Substitution)
5	arguments stack at time 20 = [ [ ], [ ] ]	if arguments stack at time (19 + 1) = arguments stack at time 20 and arguments stack at time (19 + 1) = [ [ ], [ ] ], then arguments stack at time 20 = [ [ ], [ ] ] (Transitive Property of Equality Variation 2)

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