Proof: Byte 125 Stays the Same 11

Let's prove the following theorem:

if the following are true:

instruction #17 is load dst=3 addr=1 imm=1
the PC at time 11 = 17
value of cell 125 at time 11 = 2

then value of cell 125 at time 12 = 2

Proof:

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

Given

1	instruction #17 is `load dst=3 addr=1 imm=1`
2	the PC at time 11 = 17
3	value of cell 125 at time 11 = 2

Proof Table

#	Claim	Reason
1	not (125 = 3)	not (125 = 3) (Fixed Value Statement)
2	value of cell 125 at time (11 + 1) = value of cell 125 at time 11	if instruction #17 is `load dst=3 addr=1 imm=1` and the PC at time 11 = 17 and not (125 = 3), then value of cell 125 at time (11 + 1) = value of cell 125 at time 11 (LOAD Instruction Property 2)
3	11 + 1 = 12	11 + 1 = 12 (Fixed Value Statement)
4	value of cell 125 at time (11 + 1) = value of cell 125 at time 12	if 11 + 1 = 12, then value of cell 125 at time (11 + 1) = value of cell 125 at time 12 (Substitution)
5	value of cell 125 at time 12 = value of cell 125 at time 11	if value of cell 125 at time (11 + 1) = value of cell 125 at time 12 and value of cell 125 at time (11 + 1) = value of cell 125 at time 11, then value of cell 125 at time 12 = value of cell 125 at time 11 (Transitive Property of Equality Variation 2)
6	value of cell 125 at time 12 = 2	if value of cell 125 at time 12 = value of cell 125 at time 11 and value of cell 125 at time 11 = 2, then value of cell 125 at time 12 = 2 (Transitive Property of Equality)

Comments

Please log in to add comments