MHB Halting problem is undecidable proof confusion-:

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The discussion centers on the contradiction arising from the assumption that a Turing machine H can solve the halting problem. It explains that if D halts on its own description R(D), then according to the definition of H', H' must loop on R(D). This leads to the conclusion that D cannot halt on R(D). Conversely, if D loops on R(D), then H' must halt, implying that D halts on R(D). This creates a logical paradox where both conditions cannot simultaneously hold true, resulting in the statement P (D halts on R(D)) being equivalent to its negation, leading to the conclusion that such a Turing machine H cannot exist. This contradiction underscores the undecidability of the halting problem.
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https://slideplayer.com/slide/10708471/
This is the context I am talking about.

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What contradiction occur here? We begin by telling that there is a Turing machine H that solves the halting problem. So how does this contradicts? Can you tell me about that?

What contradiction occur here? We begin by telling that there is a Turing machine H that solves the halting problem. So how does this contradicts? Can you tell me about that?
 
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I write $M(w)$ for the computation of a TM $M$ on inout $w$. The contradiction is as follows. If $D$ halts on $R(D)$, then by the definition of $H'$ we have $H'(R(D), R(D))$ loops. But the latter computation is a part of the computation $D(R(D))$; therefore, $D$ loops on $R(D)$. This is by itself is not a contradiction yet; it just means that $D$ cannot halt on $R(D)$. However, the inversion of that implication also holds: if $D$ loops on $R(D)$, then by the definition of $H'$ we have $H'(R(D), R(D))$ halts. Since this computation is the last portion of $D(R(D))$, the latter computation halts as well. So if we denote "$D$ halts on $R(D)$" by $P$, we get both $P\to\neg P$ (which implies $\neg P$) and $\neg P\to P$, which together with the former implication means $P\leftrightarrow \neg P$. This is a contradiction.
 
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