$\text{Definition:}$ Let $w_M$ be a string that describes a Turing machine $M = (Q,\Sigma,\Gamma,\delta,q_0,\square,F)$, and let $w$ be a string in $M’$s alphabet. We will assume that $w_m$ and $w_M$ and $w$ are encoded as a string of $0’$s and $1’$s. A solution of the halting problem is a Turing machine $H,$ which for any $w_M$ and $w$ performs the computation
$q_0w_Mw\vdash{^\ast} \space x_1q_yx_2$
If $M$ applied to $w$ halts, and
$q_0w_Mw\vdash{^\ast} \space y_1q_ny_2$
If $M$ applied to $w$ does not halt. Here $q_y$ and $q_n$ are both final states of $H$
$\text{Theorem:}$ There does not exist any Turing machine $H$ that behave as required by Definition. The halting problem is therefore undecidable.
Suppose we change Definition to require that $q_0w_Mw\vdash{^\ast} \space q_yw$ or $q_0w_Mw\vdash{^\ast} \space q_nw$, depending on whether $M$ applied to $w$ halts or not. Reexamine the proof of Theorem to show that this difference in the definition does not affect the proof in any significant way.
