We also know that \(a \lt n\) and \(b \lt n\text{,}\) so by our inductive hypothesis, \(P(a)\) and \(P(b)\) are true. To reduce the \(a\)-square bar to single squares takes \(a-1\) breaks; to reduce the \(b\)-square bar to single squares takes \(b-1\) breaks. Doing this results in our original bar being reduced to single squares. All together it took the initial break, plus the \(a-1\) and \(b-1\) breaks, for a total of \(1+a-1+b-1 = a+b-1 = n-1\) breaks. Thus \(P(n)\) is true.
in-context