Direct proof:
"\\text{If}\\ x\\in A\\cap B\\implies x\\in A\\implies x\\in A \\ \\text{or} \\ x\\in B \\implies""\\implies x\\in A\\cup B"
Therefore
"A\\cap B\\sube A\\cup B"
Proof by contradiction
Suppose to the contrary that "A\\cap B \\subsetneq A\\cup B"
Then an element "x\\in A\\cap B" exists such that "x\\notin A\\cup B."
That is, there is an element x that belongs to both A and B and ("x\\in A" and "x\\in B" ) at the same time belongs to neither A nor B ("x\\notin A" and "x\\notin B" ). This is a contradiction, so the original assumption is false. It follows that
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