Proof: Byte 1 Stays the Same 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
value of cell 1 at time 15 = 124

then value of cell 1 at time 16 = 124

Proof:

View as a tree | View dependent proofs | Try proving it

Given

1	instruction #2 is `load dst=3 addr=1 imm=1`
2	the PC at time 15 = 2
3	value of cell 1 at time 15 = 124

Proof Table

#	Claim	Reason
1	not (1 = 3)	not (1 = 3) (Fixed Value Statement)
2	value of cell 1 at time (15 + 1) = value of cell 1 at time 15	if instruction #2 is `load dst=3 addr=1 imm=1` and the PC at time 15 = 2 and not (1 = 3), then value of cell 1 at time (15 + 1) = value of cell 1 at time 15 (LOAD Instruction Property 2)
3	15 + 1 = 16	15 + 1 = 16 (Fixed Value Statement)
4	value of cell 1 at time (15 + 1) = value of cell 1 at time 16	if 15 + 1 = 16, then value of cell 1 at time (15 + 1) = value of cell 1 at time 16 (Substitution)
5	value of cell 1 at time 16 = value of cell 1 at time 15	if value of cell 1 at time (15 + 1) = value of cell 1 at time 16 and value of cell 1 at time (15 + 1) = value of cell 1 at time 15, then value of cell 1 at time 16 = value of cell 1 at time 15 (Transitive Property of Equality Variation 2)
6	value of cell 1 at time 16 = 124	if value of cell 1 at time 16 = value of cell 1 at time 15 and value of cell 1 at time 15 = 124, then value of cell 1 at time 16 = 124 (Transitive Property of Equality)

Comments

Please log in to add comments