Specification for the gas fee of System operations

[cuongquangnam]

In the execution model, the remaining gas after executing an operation is defined as:

And then in appendix H we have:

and in the definition for CALL opcode, we have

However, there is also definition for the cost below:

Therefore, is it that
${\mu'}_g = {\mu}_g - C_{CALLGAS}({\sigma},{\mu}, A) + g' = {\mu}_g - C({\sigma},{\mu}, A,I)= {\mu}_g - C_{mem}({\mu_{i}'}) + C_{mem}({\mu_{i}}) - C_{CALL}({\sigma},{\mu}, A)$
I really doubt it is the case, can anyone help explain this?
For me, the more intuitive equation should be:
${\mu'}_g = {\mu}_g - C_{mem}({\mu_{i}'}) + C_{mem}({\mu_{i}}) - C_{CALL}({\sigma},{\mu}, A) + g' = {\mu}_g - C({\sigma},{\mu}, A,I) + g'$
and
$C_{CALL}({\sigma, \mu, A }) = C_{CALLGAS}({\sigma, \mu, A }) + C_{EXTRA}({\sigma, \mu, A })$
Therefore, it seems that for the case of system operation (e.g. CALL), we should add g' to ${\mu'_g}$ (to take back the gas remaining after the execution of that operation)

[kodyfanz]

First:

$\boldsymbol{\mu}'_g \equiv \boldsymbol{\mu}_g - C(\boldsymbol{\sigma}, \boldsymbol{\mu}, A, I)$

$C(\boldsymbol{\sigma}, \boldsymbol{\mu}, A, I)$ in this formula represents a general description for cases involving gas costs, depending on the world-state $\boldsymbol{\sigma}$, machine-state $\boldsymbol{\mu}$, the accrued substate $A$, and the instruction $I$.

$C(\boldsymbol{\sigma},\boldsymbol{\mu},A,I) \equiv C_{mem}(\boldsymbol{\mu}_ i')-C_{mem}(\boldsymbol{\mu}_ i)+C_{\text{CALL}}(\boldsymbol{\sigma},\boldsymbol{\mu},A)$ if $w \in W_{call}$

The above formula is specified in the case where the instruction $I$ belongs to the set of CALL instructions, again:

$C_{\text{CALL}}(\boldsymbol{\sigma}, \boldsymbol{\mu}, A) \equiv C_{\text{CALLGAS}}(\boldsymbol{\sigma},\boldsymbol{\mu}, A) - g'$

where $g'$ is the amount of gas refunded after the execution of the CALL instruction.

Thus, there is no contradiction in the yellow paper here.