My answer is similar to the one of Dave Tweed, meaning that I put it on a more formal level. I obviously answered later, but I decided to nevertheless post it since someone may find this approach interesting.
The relation you are trying to prove is independent from the structure of the function f since it is, as a matter of fact, a tautology. To explain what I mean, I propose a demonstration for a general, correctly formed, Boolean expression P in an arbitrary number of Boolean variables, say n∈N, y1,…,yn, where yi∈{0,1} for all i=1,…,n.
We have that P(y1,…,yn)∈{0,1} and consider the following two sets of Boolean values for the n-dimensional Boolean vector (y1,…,yn)
YY¯={(y1,…,yn)∈{0,1}n|P(y1,…,yn)=1}={(y1,…,yn)∈{0,1}n|P(y1,…,yn)=0}
These set are a partition of the full set of values the input Boolean vector can assume, i.e. Y∪Y¯={0,1}n and Y∩Y¯=∅ (the empty set), thus
P(y1,…,yn)P′(y1,…,yn)={01if (y1,…,yn)∈Y¯if (y1,…,yn)∈Y⇕={10if (y1,…,yn)∈Y¯if (y1,…,yn)∈Y
therefore we always have
P+P′=1∀(y1,…,yn)∈{0,1}n