Proof: Store 0

Let's prove the following theorem:

if the following are true:
  • instruction #0 is store src=3 addr=6 imm=0
  • the PC at time 0 = 0
  • value of cell 6 at time 0 = 7
  • value of cell 3 at time 0 = 13

then value of cell 7 at time 1 = 13

Proof:

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

Given
1 instruction #0 is store src=3 addr=6 imm=0
2 the PC at time 0 = 0
3 value of cell 6 at time 0 = 7
4 value of cell 3 at time 0 = 13
Proof Table
# Claim Reason
1 value of cell ((value of cell 6 at time 0) + 0) at time (0 + 1) = value of cell 3 at time 0 if instruction #0 is store src=3 addr=6 imm=0 and the PC at time 0 = 0, then value of cell ((value of cell 6 at time 0) + 0) at time (0 + 1) = value of cell 3 at time 0
2 (value of cell 6 at time 0) + 0 = 7 + 0 if value of cell 6 at time 0 = 7, then (value of cell 6 at time 0) + 0 = 7 + 0
3 7 + 0 = 7 7 + 0 = 7
4 (value of cell 6 at time 0) + 0 = 7 if (value of cell 6 at time 0) + 0 = 7 + 0 and 7 + 0 = 7, then (value of cell 6 at time 0) + 0 = 7
5 value of cell ((value of cell 6 at time 0) + 0) at time (0 + 1) = value of cell 7 at time (0 + 1) if (value of cell 6 at time 0) + 0 = 7, then value of cell ((value of cell 6 at time 0) + 0) at time (0 + 1) = value of cell 7 at time (0 + 1)
6 0 + 1 = 1 0 + 1 = 1
7 value of cell 7 at time (0 + 1) = value of cell 7 at time 1 if 0 + 1 = 1, then value of cell 7 at time (0 + 1) = value of cell 7 at time 1
8 value of cell 7 at time 1 = value of cell 3 at time 0 if value of cell ((value of cell 6 at time 0) + 0) at time (0 + 1) = value of cell 7 at time (0 + 1) and value of cell 7 at time (0 + 1) = value of cell 7 at time 1 and value of cell ((value of cell 6 at time 0) + 0) at time (0 + 1) = value of cell 3 at time 0, then value of cell 7 at time 1 = value of cell 3 at time 0
9 value of cell 7 at time 1 = 13 if value of cell 7 at time 1 = value of cell 3 at time 0 and value of cell 3 at time 0 = 13, then value of cell 7 at time 1 = 13

Comments

Please log in to add comments