Functional Dependencies
$R$ has only five non-trivial functional dependencies (FDs).

No, $R$ has many more non-trivial FDs. For example, Since A-&gt; BC and C -&gt; F, we also  have AC-&gt;BF, which is also a non-trivial FD although not a irreducible FD.
What you can say is that all given 5 FDs are non-trivial FDs, since for each given FD, the dependent set is not a subset of the the determinant set.
If you want to see all non-trivial function dependencies of $R$, you have to obtain the dependency closure of the given FDs, following, for example, the procedure given here using Armstrong's axioms.

Step-1 : Add the attributes which are present on Left Hand Side in the original functional dependency.
Step-2 : Now, add the attributes present on the Right Hand Side of the functional dependency.
Step-3 : With the help of attributes present on Right Hand Side, check the other attributes that can be derived from the other given functional dependencies. Repeat this process until all the possible attributes which can be derived are added in the closure.

You can enter R(A,B,C,D,E,F) and A -&gt; B,C; A,D -&gt; B,E; C -&gt;F ; A-&gt;F; D-&gt; E into the first two fields on this page. Hit the &quot;Calculate&quot; button and view the result. It will list a lot of related concepts. In particular, it shows the following 43 non-trivial FDs.
A → B,C,F
C → F
D → E
A,B → C,F
A,C → B,F
A,D → B,C,E,F
A,E → B,C,F
A,F → B,C
B,C → F
B,D → E
C,D → E,F
C,E → F
D,F → E
A,B,C → F
A,B,D → C,E,F
A,B,E → C,F
A,B,F → C
A,C,D → B,E,F
A,C,E → B,F
A,C,F → B
A,D,E → B,C,F
A,D,F → B,C,E
A,E,F → B,C
B,C,D → E,F
B,C,E → F
B,D,F → E
C,D,E → F
C,D,F → E
A,B,C,D → E,F
A,B,C,E → F
A,B,D,E → C,F
A,B,D,F → C,E
A,B,E,F → C
A,C,D,E → B,F
A,C,D,F → B,E
A,C,E,F → B
A,D,E,F → B,C
B,C,D,E → F
B,C,D,F → E
A,B,C,D,E → F
A,B,C,D,F → E
A,B,D,E,F → C
A,C,D,E,F → B