Proof: Pc 7

Let's prove the following theorem:

if the following are true:

instruction #0 is load dst=7 addr=5 imm=0
the PC at time 7 = 0

then the PC at time 8 = 1

Proof:

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Given

1	instruction #0 is `load dst=7 addr=5 imm=0`
2	the PC at time 7 = 0

Proof Table

#	Claim	Reason
1	the PC at time (7 + 1) = 0 + 1	if instruction #0 is `load dst=7 addr=5 imm=0` and the PC at time 7 = 0, then the PC at time (7 + 1) = 0 + 1 (LOAD Instruction Property 3)
2	7 + 1 = 8	7 + 1 = 8 (Fixed Value Statement)
3	0 + 1 = 1	0 + 1 = 1 (Additive Identity Variation)
4	the PC at time (7 + 1) = the PC at time 8	if 7 + 1 = 8, then the PC at time (7 + 1) = the PC at time 8 (Substitution)
5	the PC at time 8 = 1	if the PC at time (7 + 1) = 0 + 1 and the PC at time (7 + 1) = the PC at time 8 and 0 + 1 = 1, then the PC at time 8 = 1 (Transitive Property Application 2)

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