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Question
Prove that `[bar"a" bar"b" + bar"c" bar"a" + bar"b" + bar"c"] = 0`
Solution
`[bar"a" bar"b" + bar"c" bar"a" + bar"b" + bar"c"]`
`= bar"a" * [(bar"b" + bar"c") xx (bar"a" + bar"b" + bar"c")]`
`= bar"a" * (bar"b" xx bar"a" + bar"b" xx bar"b" + bar"b" xx bar"c" + bar"c" xx bar"a" + bar"c" xx bar"b" + bar"c" xx bar"c")`
`= bar"a" * (bar"b" xx bar"a") + bar"a" * (bar"b" xx bar"b") + bar"a" * (bar"b" xx bar"c") + bar"a" * (bar"c" xx bar"a") + bar"a" * (bar"c" xx bar"b") + bar"a" * (bar"c" xx bar"c")`
`= 0 + 0 + bar"a" * (bar"b" xx bar"c") + 0 - bar"a" * (bar"b" xx bar"c") + 0`
= 0
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