COSMAC ELF 1802 commands?

Question

assembler

The list of 1802 teams is here .
I am disassembling the lamp blinking program on the COSMAC ELF emulator from the site . First comes command 7A - turn off the light bulb. The command loads the number 2 into register R0. Why?
Next comes the command F8 Description of this command M(R(P)) → D; R(P) + 1 → R(P)
The F8 command loads the number 3 into the F8 register, although the command, judging by the description, must add the number 1 to the R0 register and store the result in R0, which means that the number 1 should appear in R0, not so whether?
Command 7B - turn on the lamp - also register R0 number 2.

The entire program for flashing the lamp
7A
F8
10
B1
21
91
3A
04
31
00
7B
30
01
00

Answer 1

First comes command 7A - turn off the light bulb. The command loads the number 2 into register R0. Why?

Since you have P = 0, then R(0) is used as a program counter. Instruction 7A should increase it by 1, i.e. after it, R(0) must be 1.

Next comes the command F8 Description of this command M(R(P)) → D; R(P) + 1 → R(P)
The F8 command loads the number 3 into the F8 register, although the command, judging by the description, must add the number 1 to the R0 register and store the result in R0, which means that the number 1 should end up in R0.

No not like this. First, the instruction is loaded into registers I and N, and R(0) is incremented by 1. After that, a byte is taken from memory at address R(0) and placed in D, and R(0) is incremented by another 1.
The disassembled program:

0: 7A -- reset Q (Q = 0)
1: F8 -- ldi 10 (D = 10)
2: 10
3: B1 -- phi 1 (R(1).1 = D, т.е. R(1) = 1000)
4: 21 -- dec 1 (R(1) = R(1) - 1, т.е. R(1) = fff)
5: 91 -- ghi 1 (D = R(1).1, т.е. D = f)
6: 3A -- bnz 4 (переход к адресу 4, пока D != 0)
7: 04
8: 31 -- bq 0 (переход к адресу 0 если Q == 1)
9: 00
a: 7B -- set Q (Q = 1)
b: 30 -- br 1 (переход к адресу 1)
с: 01
d: 00 -- idle (никогда не попадаем сюда).

In the middle, just a delay loop on the counter.