Proof: Byte 7 Stays the Same 1

Let's prove the following theorem:

if the following are true:

instruction #1 is load dst=8 addr=6 imm=1
the PC at time 1 = 1
value of cell 7 at time 1 = 13

then value of cell 7 at time 2 = 13

Proof:

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

Given

1	instruction #1 is `load dst=8 addr=6 imm=1`
2	the PC at time 1 = 1
3	value of cell 7 at time 1 = 13

Proof Table

#	Claim	Reason
1	not (7 = 8)	not (7 = 8) (Fixed Value Statement)
2	value of cell 7 at time (1 + 1) = value of cell 7 at time 1	if instruction #1 is `load dst=8 addr=6 imm=1` and the PC at time 1 = 1 and not (7 = 8), then value of cell 7 at time (1 + 1) = value of cell 7 at time 1 (LOAD Instruction Property 2)
3	1 + 1 = 2	1 + 1 = 2 (Fixed Value Statement)
4	value of cell 7 at time (1 + 1) = value of cell 7 at time 2	if 1 + 1 = 2, then value of cell 7 at time (1 + 1) = value of cell 7 at time 2 (Substitution)
5	value of cell 7 at time 2 = value of cell 7 at time 1	if value of cell 7 at time (1 + 1) = value of cell 7 at time 2 and value of cell 7 at time (1 + 1) = value of cell 7 at time 1, then value of cell 7 at time 2 = value of cell 7 at time 1 (Transitive Property of Equality Variation 2)
6	value of cell 7 at time 2 = 13	if value of cell 7 at time 2 = value of cell 7 at time 1 and value of cell 7 at time 1 = 13, then value of cell 7 at time 2 = 13 (Transitive Property of Equality)

Comments

Please log in to add comments