SC2056

if  (( $1 != 0 || $1 != 3 ))
then
  echo "$1 is not 0 or 3"
fi

if  (( $1 != 0 && $1 != 3 ))
then
  echo "$1 is not 0 or 3"
fi

This is not a bash issue, but a simple, common logical mistake applicable to all languages.

(( $1 != 0 || $1 != 3 )) is always true:

• If $1 = 0 then $1 != 3 is true, so the statement is true.
• If $1 = 3 then $1 != 0 is true, so the statement is true.
• If $1 = 42 then $1 != 0 is true, so the statement is true.

(( $1 != 0 && $1 != 3 )) is true only when $1 is not 0 and not 3:

• If $1 = 0, then $1 != 3 is false, so the statement is false.
• If $1 = 3, then $1 != 0 is false, so the statement is false.
• If $1 = 42, then both $1 != 0 and $1 != 3 is true, so the statement is true.

This statement is identical to ! (( $1 == 0 || $1 == 3 )), which also works correctly.

None.