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+## You probably wanted && here, otherwise it's always true.
+
+### Problematic code:
+
+```sh
+if [ "$1" != foo ] || [ "$1" != bar ]
+then
+  echo "$1 is not foo or bar"
+fi
+```
+
+### Correct code:
+
+```sh
+if [ "$1" != foo ] && [ "$1" != bar ]
+then
+  echo "$1 is not foo or bar"
+fi
+```
+
+### Rationale:
+
+This is not a bash issue, but a simple, common logical mistake applicable to all languages. 
+
+`[ "$1" != foo ] || [ "$1" != bar ]` is always true (when `foo != bar`):
+
+* If `$1 = foo` then `$1 != bar` is true, so the statement is true.
+* If `$1 = bar` then `$1 != foo` is true, so the statement is true.
+* If `$1 = cow` then `$1 != foo` is true, so the statement is true.
+
+`[ $1 != foo ] && [ $1 != bar ]` matches when `$1` is neither `foo` nor `bar`:
+
+* If `$1 = foo`, then `$1 != foo` is false, so the statement is false.
+* If `$1 = bar`, then `$1 != bar` is false, so the statement is false.
+* If `$1 = cow`, then both `$1 != foo` and `$1  != bar` is true, so the statement is true.
+
+This statement is identical to `! [ "$1" = foo ] || [ "$1" = bar ]`, which also works correctly (by [De Morgan's law](https://en.wikipedia.org/wiki/De_Morgan%27s_laws))
+
+This warning is equivalent to [[SC2055]] and [[SC2056]], which trigger for intra-`test` expressions and arithmetic contexts respectively.
+
+### Exceptions
+
+Rare.
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