Proof: Load 0

Let's prove the following theorem:

if the following are true:
  • instruction #0 is load dst=7 addr=5 imm=0
  • the PC at time 0 = 0
  • value of cell 5 at time 0 = 8
  • value of cell 8 at time 0 = 24

then value of cell 7 at time 1 = 24

Proof:

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Given
1 instruction #0 is load dst=7 addr=5 imm=0
2 the PC at time 0 = 0
3 value of cell 5 at time 0 = 8
4 value of cell 8 at time 0 = 24
Proof Table
# Claim Reason
1 value of cell 7 at time (0 + 1) = value of cell ((value of cell 5 at time 0) + 0) at time 0 if instruction #0 is load dst=7 addr=5 imm=0 and the PC at time 0 = 0, then value of cell 7 at time (0 + 1) = value of cell ((value of cell 5 at time 0) + 0) at time 0
2 (value of cell 5 at time 0) + 0 = 8 + 0 if value of cell 5 at time 0 = 8, then (value of cell 5 at time 0) + 0 = 8 + 0
3 8 + 0 = 8 8 + 0 = 8
4 (value of cell 5 at time 0) + 0 = 8 if (value of cell 5 at time 0) + 0 = 8 + 0 and 8 + 0 = 8, then (value of cell 5 at time 0) + 0 = 8
5 value of cell ((value of cell 5 at time 0) + 0) at time 0 = value of cell 8 at time 0 if (value of cell 5 at time 0) + 0 = 8, then value of cell ((value of cell 5 at time 0) + 0) at time 0 = value of cell 8 at time 0
6 value of cell ((value of cell 5 at time 0) + 0) at time 0 = 24 if value of cell ((value of cell 5 at time 0) + 0) at time 0 = value of cell 8 at time 0 and value of cell 8 at time 0 = 24, then value of cell ((value of cell 5 at time 0) + 0) at time 0 = 24
7 value of cell 7 at time (0 + 1) = 24 if value of cell 7 at time (0 + 1) = value of cell ((value of cell 5 at time 0) + 0) at time 0 and value of cell ((value of cell 5 at time 0) + 0) at time 0 = 24, then value of cell 7 at time (0 + 1) = 24
8 0 + 1 = 1 0 + 1 = 1
9 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
10 value of cell 7 at time 1 = 24 if value of cell 7 at time (0 + 1) = value of cell 7 at time 1 and value of cell 7 at time (0 + 1) = 24, then value of cell 7 at time 1 = 24

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