Boolean SOP Expression Simplification: $F(a,b,c,d) = (a+d)(a'b+c'd)(ac+bd)'$

Question

my answer that I have gotten is $b'c'd + a' b d'$ however, the answer given to me was b'c'd can someone tell me whether I am correct

Answer 1

I calculated, the correct answer is b'c'd

Could you show the steps you used?

Answer 2

1530261   $F\equiv((a\lor d)\land((\lnot a\land b)\lor(\lnot c\land d))\land\lnot((a\land c)\lor(b\land d)))$

$\begin{array} ee\equiv(a\lor d)\\ f\equiv(\lnot a)\\ g\equiv(f\land b)\\ h\equiv(\lnot c)\\ i\equiv(h\land d)\\ j\equiv(g\lor i)\\ k\equiv(e\land j)\\ l\equiv(a\land c)\\ m\equiv(b\land d)\\ n\equiv(l\lor m)\\ o\equiv(\lnot n)\\ \hline F\equiv(k\land o) \end{array}$

$\begin{array}{cccc|ccccccccccc|c} a&b&c&d&e&f&g&h&i&j&k&l&m&n&o&F\\ \hline 0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0\\ 0&0&0&1&1&1&0&1&1&1&1&0&0&0&1&1\\ 0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0\\ 0&0&1&1&1&1&0&0&0&0&0&0&0&0&1&0\\ 0&1&0&0&0&1&1&1&0&1&0&0&0&0&1&0\\ 0&1&0&1&1&1&1&1&1&1&1&0&1&1&0&0\\ 0&1&1&0&0&1&1&0&0&1&0&0&0&0&1&0\\ 0&1&1&1&1&1&1&0&0&1&1&0&1&1&0&0\\ 1&0&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\ 1&0&0&1&1&0&0&1&1&1&1&0&0&0&1&1\\ 1&0&1&0&1&0&0&0&0&0&0&1&0&1&0&0\\ 1&0&1&1&1&0&0&0&0&0&0&1&0&1&0&0\\ 1&1&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\ 1&1&0&1&1&0&0&1&1&1&1&0&1&1&0&0\\ 1&1&1&0&1&0&0&0&0&0&0&1&0&1&0&0\\ 1&1&1&1&1&0&0&0&0&0&0&1&1&1&0&0 \end{array}$

This is a step-by-step truth table of your logical function. As you can see, somewhere in your parsing you came up with an extra term.