Proof: Byte 1 Stays the Same 2

Let's prove the following theorem:

if the following are true:

instruction #2 is addi dst=4 src=0 imm=0
the PC at time 2 = 2
value of cell 1 at time 2 = 0

then value of cell 1 at time 3 = 0

Instructions

addi dst=4 src=0 imm=0

Memory Cells
0
1
2
3
4

Program Counter	Time
2	2

LW Computer Simulator

Proof:

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

Given

1	instruction #2 is `addi dst=4 src=0 imm=0`
2	the PC at time 2 = 2
3	value of cell 1 at time 2 = 0

Proof Table

#	Claim	Reason
1	not (1 = 4)	not (1 = 4) (Fixed Value Statement)
2	value of cell 1 at time (2 + 1) = value of cell 1 at time 2	if instruction #2 is `addi dst=4 src=0 imm=0` and the PC at time 2 = 2 and not (1 = 4), then value of cell 1 at time (2 + 1) = value of cell 1 at time 2 (ADDI Instruction Property 2)
3	2 + 1 = 3	2 + 1 = 3 (Fixed Value Statement)
4	value of cell 1 at time (2 + 1) = value of cell 1 at time 3	if 2 + 1 = 3, then value of cell 1 at time (2 + 1) = value of cell 1 at time 3 (Substitution)
5	value of cell 1 at time 3 = value of cell 1 at time 2	if value of cell 1 at time (2 + 1) = value of cell 1 at time 3 and value of cell 1 at time (2 + 1) = value of cell 1 at time 2, then value of cell 1 at time 3 = value of cell 1 at time 2 (Transitive Property of Equality Variation 2)
6	value of cell 1 at time 3 = 0	if value of cell 1 at time 3 = value of cell 1 at time 2 and value of cell 1 at time 2 = 0, then value of cell 1 at time 3 = 0 (Transitive Property of Equality)

Comments

Please log in to add comments