My binary logic for my circuit is \begin{array}{|c|c|c|c|c|c|} A &B &C &D &Hallway &Stairs\\\hline 0 &0 &0 &0 &0 &0\\ 0 &0 &0 &1 &1 &1\\ 0 &0 &1 &0 &1 &0\\ 0 &0 &1 &1 &X &X\\ 0 &1 &0 &0 &1 &0\\ 0 &1 &0 &1 &X &X\\ 0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &X &X\\ 1 &0 &0 &0 &1 &1\\ 1 &0 &0 &1 &0 &0\\ 1 &0 &1 &0 &X &X\\ 1 &0 &1 &1 &X &X\\ 1 &1 &0 &0 &X &X\\ 1 &1 &0 &1 &X &X\\ 1 &1 &1 &0 &X &X\\ 1 &1 &1 &1 &X &X\\\hline \end{array}
The expression for the hallway is$$ \bar A\bar B\bar C D + \bar A\bar BC\bar D + \bar A B \bar C\bar D + A\bar B \bar C\bar D $$ simplified after a week of brute force guessing is $$ \bar A D + \bar BC + B\bar C+ A\bar D $$
Karnaugh map turned out \begin{array}{|c|c|c|c|c|}\hline \large CD\backslash AB\normalsize &00 &01 &11 &10 \\\hline 00 &0 &1 &0 &1 \\ 01 &1 &0 &0 &0 \\ 11 &0 &0 &0 &0 \\ 10 &1 &0 &0 &0 \\\hline \end{array}
The reason why I couldn't come to a simplification normally, is I kept coming to the snipet $ \bar A \bar D (B \bar C + \bar BC )$ minus the $\bar A \bar D $ , the 2 expressions in parenthesis is a XOR gate. Which I then tried substituting that for $ \bar A \bar D (B + C) (\bar B + \bar C). $ But didn't help me much. This was apparent in a question answered very blantantly from Mauro ALLEGRANZA "No guy, you can't"
Question: Explain to me as if I were a beginner, on how to simplify $ \bar A\bar B\bar C D + \bar A\bar BC\bar D + \bar A B \bar C\bar D + A\bar B \bar C\bar D $. Adding what Boolean Theorems you used. Please and thanks.
The simple answer is: you don't.
The expression $\bar A\bar B\bar CD + \bar A\bar BC\bar D + \bar AB\bar C\bar D + A\bar B\bar C\bar D$ is already as simple as it can get in disjunctive normal form. Other standard forms are not significantly better.
You could write it as $\bar A\bar B(\bar C D + C\bar D) + (\bar A B + A\bar B) \bar C\bar D$.
$\bar A\bar B\bar CD + \bar A\bar BC\bar D + \bar AB\bar C\bar D + A\bar B\bar C\bar D$ is not equivalent to $\bar A D + \bar BC + B\bar C+ A\bar D.$ There are numerous counterexamples; for example, if $A=0$ and $B=C=D=1$ then $\bar A D + \bar BC + B\bar C+ A\bar D$ is true but $\bar A\bar B\bar CD + \bar A\bar BC\bar D + \bar AB\bar C\bar D + A\bar B\bar C\bar D$ is false.
The reason you can express your necessary hallway logic as $\bar A D + \bar BC + B\bar C+ A\bar D$ is because $\bar A\bar B\bar CD + \bar A\bar BC\bar D + \bar AB\bar C\bar D + A\bar B\bar C\bar D$ is not your hallway logic. The expression $\bar A\bar B\bar CD + \bar A\bar BC\bar D + \bar AB\bar C\bar D + A\bar B\bar C\bar D$ is what your hallway logic would be if all the $X$s in the first table in the question were changed to $0$s.
That's basically what happened in the Karnaugh map in the question. You copied the $1$s faithfully from the first table, you copied the $0$s faithfully from the first table, and then you just filled everything else in with a $0$ instead of an $X$.
A Karnaugh map that accurately represented your original circuit logic (without forcing the $X$ values to be something they don't need to be) would be this:
\begin{array}{|c|c|c|c|c|}\hline \large CD\backslash AB\normalsize &00 &01 &11 &10 \\\hline 00 &0 &1 &X &1 \\ 01 &1 &X &X &0 \\ 11 &X &X &X &X \\ 10 &1 &0 &X &X \\\hline \end{array}
That should make it a lot easier to come up with your simplified formula without having to spend a week guessing.