expr_trans_bool() performs an incorrect transformation.

[Test Code]

    config MODULES
            def_bool y
            modules

    config A
            def_bool y
            select C if B != n

    config B
            def_tristate m

    config C
            tristate

[Result]

    CONFIG_MODULES=y
    CONFIG_A=y
    CONFIG_B=m
    CONFIG_C=m

This output is incorrect because CONFIG_C=y is expected.

Documentation/kbuild/kconfig-language.rst clearly explains the function
of the '!=' operator:

    If the values of both symbols are equal, it returns 'n',
    otherwise 'y'.

Therefore, the statement:

    select C if B != n

should be equivalent to:

    select C if y

Or, more simply:

    select C

Hence, the symbol C should be selected by the value of A, which is 'y'.

However, expr_trans_bool() wrongly transforms it to:

    select C if B

Therefore, the symbol C is selected by (A && B), which is 'm'.

The comment block of expr_trans_bool() correctly explains its intention:

  * bool FOO!=n => FOO
    ^^^^

If FOO is bool, FOO!=n can be simplified into FOO. This is correct.

However, the actual code performs this transformation when FOO is
tristate:

    if (e->left.sym->type == S_TRISTATE) {
                             ^^^^^^^^^^

While it can be fixed to S_BOOLEAN, there is no point in doing so
because expr_tranform() already transforms FOO!=n to FOO when FOO is
bool. (see the "case E_UNEQUAL" part)

expr_trans_bool() is wrong and unnecessary.
Signed-off-by: Masahiro Yamada <masahiroy@kernel.org>
Acked-by: Randy Dunlap <rdunlap@infradead.org>