Proof: Pc 15

Let's prove the following theorem:

if the following are true:

instruction #2 is load dst=3 addr=1 imm=1
the PC at time 15 = 2

then the PC at time 16 = 3

Proof:

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Given

1	instruction #2 is `load dst=3 addr=1 imm=1`
2	the PC at time 15 = 2

Proof Table

#	Claim	Reason
1	the PC at time (15 + 1) = 2 + 1	if instruction #2 is `load dst=3 addr=1 imm=1` and the PC at time 15 = 2, then the PC at time (15 + 1) = 2 + 1 (LOAD Instruction Property 3)
2	15 + 1 = 16	15 + 1 = 16 (Fixed Value Statement)
3	2 + 1 = 3	2 + 1 = 3 (Fixed Value Statement)
4	the PC at time (15 + 1) = the PC at time 16	if 15 + 1 = 16, then the PC at time (15 + 1) = the PC at time 16 (Substitution)
5	the PC at time 16 = 3	if the PC at time (15 + 1) = 2 + 1 and the PC at time (15 + 1) = the PC at time 16 and 2 + 1 = 3, then the PC at time 16 = 3 (Transitive Property Application 2)

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