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We will show that (A -> B) v (B -> A).
To that end, we will assume ~ ((A -> B) v (B -> A)) and show _|_.
So let us assume that ~ ((A -> B) v (B -> A)).
Then, by the first de Morgan's law, we get ~ (A -> B) & ~ (B -> A) to prove _|_.
So by splitting we get ~ (A -> B), ~ (B -> A) to prove _|_.
By implication negation, we get A, ~ (B -> A) to prove _|_.
By implication negation, we get A, ~ A to prove _|_.
Hence we get _|_ as desired.
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