Proof: Byte 126 Stays the Same 8

Let's prove the following theorem:

if the following are true:

instruction #14 is load dst=3 addr=1 imm=1
the PC at time 8 = 14
value of cell 126 at time 8 = 3

then value of cell 126 at time 9 = 3

Proof:

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Given

1	instruction #14 is `load dst=3 addr=1 imm=1`
2	the PC at time 8 = 14
3	value of cell 126 at time 8 = 3

Proof Table

#	Claim	Reason
1	not (126 = 3)	not (126 = 3) (Fixed Value Statement)
2	value of cell 126 at time (8 + 1) = value of cell 126 at time 8	if instruction #14 is `load dst=3 addr=1 imm=1` and the PC at time 8 = 14 and not (126 = 3), then value of cell 126 at time (8 + 1) = value of cell 126 at time 8 (LOAD Instruction Property 2)
3	8 + 1 = 9	8 + 1 = 9 (Fixed Value Statement)
4	value of cell 126 at time (8 + 1) = value of cell 126 at time 9	if 8 + 1 = 9, then value of cell 126 at time (8 + 1) = value of cell 126 at time 9 (Substitution)
5	value of cell 126 at time 9 = value of cell 126 at time 8	if value of cell 126 at time (8 + 1) = value of cell 126 at time 9 and value of cell 126 at time (8 + 1) = value of cell 126 at time 8, then value of cell 126 at time 9 = value of cell 126 at time 8 (Transitive Property of Equality Variation 2)
6	value of cell 126 at time 9 = 3	if value of cell 126 at time 9 = value of cell 126 at time 8 and value of cell 126 at time 8 = 3, then value of cell 126 at time 9 = 3 (Transitive Property of Equality)

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