Proof: LOAD Instruction Example 2

Let's prove the following theorem:

if the following are true:

instruction #0 is load dst=5 addr=1 imm=0
the PC at time 0 = 0
value of cell 1 at time 0 = 3
value of cell 3 at time 0 = 9

then value of cell 5 at time 1 = 9

The following example shows that when a LOAD instruction with dst = 5, addr = 1, and imm = 0 is executed, the computer looks up memory cell #1 and finds 3, then copies the value memory cell #3 (9) to memory cell #5. imm is 0 so the source address is unchanged.

Instructions

Memory Cells

Program Counter	Time
0	0

LW Computer Simulator

Proof:

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Given

1	instruction #0 is `load dst=5 addr=1 imm=0`
2	the PC at time 0 = 0
3	value of cell 1 at time 0 = 3
4	value of cell 3 at time 0 = 9

Proof Table

#	Claim	Reason
1	value of cell 5 at time (0 + 1) = value of cell ((value of cell 1 at time 0) + 0) at time 0	if instruction #0 is `load dst=5 addr=1 imm=0` and the PC at time 0 = 0, then value of cell 5 at time (0 + 1) = value of cell ((value of cell 1 at time 0) + 0) at time 0 (LOAD Instruction Property)
2	(value of cell 1 at time 0) + 0 = 3 + 0	if value of cell 1 at time 0 = 3, then (value of cell 1 at time 0) + 0 = 3 + 0 (Substitution)
3	3 + 0 = 3	3 + 0 = 3 (Additive Identity)
4	(value of cell 1 at time 0) + 0 = 3	if (value of cell 1 at time 0) + 0 = 3 + 0 and 3 + 0 = 3, then (value of cell 1 at time 0) + 0 = 3 (Transitive Property of Equality)
5	value of cell ((value of cell 1 at time 0) + 0) at time 0 = value of cell 3 at time 0	if (value of cell 1 at time 0) + 0 = 3, then value of cell ((value of cell 1 at time 0) + 0) at time 0 = value of cell 3 at time 0 (Substitution)
6	value of cell ((value of cell 1 at time 0) + 0) at time 0 = 9	if value of cell ((value of cell 1 at time 0) + 0) at time 0 = value of cell 3 at time 0 and value of cell 3 at time 0 = 9, then value of cell ((value of cell 1 at time 0) + 0) at time 0 = 9 (Transitive Property of Equality)
7	value of cell 5 at time (0 + 1) = 9	if value of cell 5 at time (0 + 1) = value of cell ((value of cell 1 at time 0) + 0) at time 0 and value of cell ((value of cell 1 at time 0) + 0) at time 0 = 9, then value of cell 5 at time (0 + 1) = 9 (Transitive Property of Equality)
8	0 + 1 = 1	0 + 1 = 1 (Additive Identity Variation)
9	value of cell 5 at time (0 + 1) = value of cell 5 at time 1	if 0 + 1 = 1, then value of cell 5 at time (0 + 1) = value of cell 5 at time 1 (Substitution)
10	value of cell 5 at time 1 = 9	if value of cell 5 at time (0 + 1) = value of cell 5 at time 1 and value of cell 5 at time (0 + 1) = 9, then value of cell 5 at time 1 = 9 (Transitive Property of Equality Variation 2)

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