Proof: Byte 1 Stays the Same 17

Let's prove the following theorem:

if the following are true:

instruction #4 is add dst=3 src1=3 src2=4
the PC at time 17 = 4
value of cell 1 at time 17 = 124

then value of cell 1 at time 18 = 124

Proof:

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

Given

1	instruction #4 is `add dst=3 src1=3 src2=4`
2	the PC at time 17 = 4
3	value of cell 1 at time 17 = 124

Proof Table

#	Claim	Reason
1	not (1 = 3)	not (1 = 3) (Fixed Value Statement)
2	value of cell 1 at time (17 + 1) = value of cell 1 at time 17	if instruction #4 is `add dst=3 src1=3 src2=4` and the PC at time 17 = 4 and not (1 = 3), then value of cell 1 at time (17 + 1) = value of cell 1 at time 17 (ADD Instruction Property 2)
3	17 + 1 = 18	17 + 1 = 18 (Fixed Value Statement)
4	value of cell 1 at time (17 + 1) = value of cell 1 at time 18	if 17 + 1 = 18, then value of cell 1 at time (17 + 1) = value of cell 1 at time 18 (Substitution)
5	value of cell 1 at time 18 = value of cell 1 at time 17	if value of cell 1 at time (17 + 1) = value of cell 1 at time 18 and value of cell 1 at time (17 + 1) = value of cell 1 at time 17, then value of cell 1 at time 18 = value of cell 1 at time 17 (Transitive Property of Equality Variation 2)
6	value of cell 1 at time 18 = 124	if value of cell 1 at time 18 = value of cell 1 at time 17 and value of cell 1 at time 17 = 124, then value of cell 1 at time 18 = 124 (Transitive Property of Equality)

Comments

Please log in to add comments