Proof: Load 2

Let's prove the following theorem:

if the following are true:
  • instruction #2 is load dst=9 addr=6 imm=2
  • the PC at time 2 = 2
  • value of cell 6 at time 2 = 3
  • value of cell 5 at time 2 = 20

then value of cell 9 at time 3 = 20

Proof:

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Given
1 instruction #2 is load dst=9 addr=6 imm=2
2 the PC at time 2 = 2
3 value of cell 6 at time 2 = 3
4 value of cell 5 at time 2 = 20
Proof Table
# Claim Reason
1 value of cell 9 at time (2 + 1) = value of cell ((value of cell 6 at time 2) + 2) at time 2 if instruction #2 is load dst=9 addr=6 imm=2 and the PC at time 2 = 2, then value of cell 9 at time (2 + 1) = value of cell ((value of cell 6 at time 2) + 2) at time 2
2 (value of cell 6 at time 2) + 2 = 3 + 2 if value of cell 6 at time 2 = 3, then (value of cell 6 at time 2) + 2 = 3 + 2
3 3 + 2 = 5 3 + 2 = 5
4 (value of cell 6 at time 2) + 2 = 5 if (value of cell 6 at time 2) + 2 = 3 + 2 and 3 + 2 = 5, then (value of cell 6 at time 2) + 2 = 5
5 value of cell ((value of cell 6 at time 2) + 2) at time 2 = value of cell 5 at time 2 if (value of cell 6 at time 2) + 2 = 5, then value of cell ((value of cell 6 at time 2) + 2) at time 2 = value of cell 5 at time 2
6 value of cell ((value of cell 6 at time 2) + 2) at time 2 = 20 if value of cell ((value of cell 6 at time 2) + 2) at time 2 = value of cell 5 at time 2 and value of cell 5 at time 2 = 20, then value of cell ((value of cell 6 at time 2) + 2) at time 2 = 20
7 value of cell 9 at time (2 + 1) = 20 if value of cell 9 at time (2 + 1) = value of cell ((value of cell 6 at time 2) + 2) at time 2 and value of cell ((value of cell 6 at time 2) + 2) at time 2 = 20, then value of cell 9 at time (2 + 1) = 20
8 2 + 1 = 3 2 + 1 = 3
9 value of cell 9 at time (2 + 1) = value of cell 9 at time 3 if 2 + 1 = 3, then value of cell 9 at time (2 + 1) = value of cell 9 at time 3
10 value of cell 9 at time 3 = 20 if value of cell 9 at time (2 + 1) = value of cell 9 at time 3 and value of cell 9 at time (2 + 1) = 20, then value of cell 9 at time 3 = 20

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