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Sum
Prove that `(bar"a" xx bar"b").(bar"c" xx bar"d")` =
`|bar"a".bar"c" bar"b".bar"c"|`
`|bar"a".bar"d" bar"b".bar"d"|.`
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Solution
Let `bar"a" xx bar"b" = bar"m"`
LHS = `(bar"a" xx bar"b").(bar"c".bar"d")`
`= bar"m".(bar"c" xx bar"d")`
`= (bar"m" xx bar"c").bar"d"`
`= [(bar"a" xx bar"b") xx bar"c"].bar"d"`
`= [(bar"c".bar"a")bar"b" - (bar"c". bar"b")bar"a"].bar"d"`
`= (bar"c".bar"a")(bar"b".bar"d") - (bar"c".bar"b")(bar"a".bar"d")`
= `|bar"c".bar"a" bar"c".bar"b"|`
`|bar"a".bar"d" bar"b".bar"d"|.`
= `|bar"a".bar"c" bar"b".bar"c"|`
`|bar"a".bar"d" bar"b".bar"d"|.` .....[Do product is commutative]
= RHS.
Concept: Section Formula
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