Proof by modus tollens To prove p ⇒ q we prove [( ∼ q ) ∧ ( p ⇒ q )] ⇒ ( ∼ p )
We know, that if a right triangle has sides of lengths a, b, c , which c is the largest, then a 2 + b 2 = c 2 ( p ⇒ q ).
Therefore, if we prove that for any triangle with sides of
lengths x, y, z , where z is the largest, it is not true, that x 2 + y 2 = z 2 ( x 2 + y 2 6 = z 2), we may conclude - by
modus tollens - that this is not a right triangle.
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