23.2 Truth tables

One very powerful (but perhaps not very elegant) way to reason about propositions is by writing a big table. The advantage of this is that it’s easy to ensure that we’ve considered every case (provided that we list every possible value). To do this, we must list them systematically. The convention is to combine the approach in Section 18.3 with the constraint that we want to list all variables possibilities in increasing order of size. I realise this probably makes no sense, so it’s best to proceed with an example:

Example 23.2.1

Prove that A(AB)AA\land(A\lor B)\equiv A.

We want to draw a table which contains every possible value for AA and BB. Note that we want to list the possibilities together, rather than separately (i.e. there are 44 total values to list as there are 22 possible values for AA and then for each of those possibilities there are 22 possible values for BB)22 2 See Section 18.1.2 if you are unsure about this.. What about this weird \sayincreasing order of size? First, examine the table

AA BB ABA\lor B A(AB)A\land(A\lor B)
F F F F
F T T F
T F T T
T T T T

If we set F=0F=0 and T=1T=1, and then read the column AA and that of BB and concatenate the two values (e.g. in the first row of the table, take A=F=0A=F=0 and B=F=0B=F=0, so their concatenation is the string 0 followed by the string 0 again), then we wish to list them in increasing value (this just makes everything much clearer, and also makes it harder to make mistakes). In the table above we have 00<01<10<1100<01<10<11. The same idea extends to larger tables.

Returning to the formula, it must be true, as the column for AA is the same as the column for A(AB)A\land(A\lor B).