Autosave: 2022-12-17 16:30:05

Logic/Propositional_logic/Boolean_functions.md

@@ -15,7 +15,18 @@ $$
 Here is a work through where $f(1, 0, 1)$:
 
 - The first disjunction : $ \lnot(x) \land z $ is false because $x$ is 1 and $z$ is 1
-- The second disjunction: $x \land y$ is false because $x$ is 1 and $y$ is 0
+- The second disjunction: $x \land y$ is true because $x$ is 1 and $y$ is 1
 - The overall function returns true because the main connective is disjunction and one of the disjuncts (the second) evaluates to 1. Thus the output is 1.
 
-We can compute all possible outputs of the function by constructing a truth-table with each possible variable as the truth conditions and the output of the function as the truth value:
+We can compute all possible outputs of the function by constructing a [truth-table](/Logic/Propositional_logic/Truth-tables.md) with each possible variable as the truth conditions and the output of the function as the truth value:
+
+| $x$ | $y$ | $z$ | $f(x,y,z) = (x \land y) \lor (\lnot(x) \land z )$ |
+| --- | --- | --- | ------------------------------------------------- |
+| 0   | 0   | 0   | 0                                                 |
+| 0   | 0   | 1   | 1                                                 |
+| 0   | 1   | 0   | 0                                                 |
+| 0   | 1   | 1   | 1                                                 |
+| 1   | 0   | 0   | 0                                                 |
+| 1   | 0   | 1   | 0                                                 |
+| 1   | 1   | 0   | 1                                                 |
+| 1   | 1   | 1   | 1                                                 |