Proof: Byte 126 Stays the Same 21

Let's prove the following theorem:

if the following are true:

instruction #21 is load dst=3 addr=1 imm=0
the PC at time 21 = 21
value of cell 126 at time 21 = 3

then value of cell 126 at time 22 = 3

Proof:

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Given

1	instruction #21 is `load dst=3 addr=1 imm=0`
2	the PC at time 21 = 21
3	value of cell 126 at time 21 = 3

Proof Table

#	Claim	Reason
1	not (126 = 3)	not (126 = 3) (Fixed Value Statement)
2	value of cell 126 at time (21 + 1) = value of cell 126 at time 21	if instruction #21 is `load dst=3 addr=1 imm=0` and the PC at time 21 = 21 and not (126 = 3), then value of cell 126 at time (21 + 1) = value of cell 126 at time 21 (LOAD Instruction Property 2)
3	21 + 1 = 22	21 + 1 = 22 (Fixed Value Statement)
4	value of cell 126 at time (21 + 1) = value of cell 126 at time 22	if 21 + 1 = 22, then value of cell 126 at time (21 + 1) = value of cell 126 at time 22 (Substitution)
5	value of cell 126 at time 22 = value of cell 126 at time 21	if value of cell 126 at time (21 + 1) = value of cell 126 at time 22 and value of cell 126 at time (21 + 1) = value of cell 126 at time 21, then value of cell 126 at time 22 = value of cell 126 at time 21 (Transitive Property of Equality Variation 2)
6	value of cell 126 at time 22 = 3	if value of cell 126 at time 22 = value of cell 126 at time 21 and value of cell 126 at time 21 = 3, then value of cell 126 at time 22 = 3 (Transitive Property of Equality)

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